Explanation:
Okay, so I need to figure out the difference between the area of the region with the greatest area and the area of the region with the smallest area among the three regions labeled X, Y, and Z in the diagram. The diagram shows three concentric circles with radii 4, 6, and 7.

First, let me visualize the diagram. There are three circles with the same center, right? The innermost circle has radius 4, then the middle one has radius 6, and the outermost one has radius 7. The regions X, Y, Z are probably the areas between these circles. The Asymptote code labels X at (2,0), which is probably the area between radius 0 and 4, Y at (5,0), which might be between 4 and 6, and Z at (6.5,0), which is between 6 and 7. Wait, but the Asymptote code fills the circles with different colors: the largest circle (radius 7) is light gray, the middle one (radius 6) is gray, and the innermost (radius 4) is white. So maybe the regions are:

- Region X: the white circle (radius 4), which is the innermost area.

- Region Y: the area between radius 4 and 6, which is the gray part minus the white part?

Wait, but the Asymptote code says:

filldraw(circle((0,0),7), lightgray, black+linewidth(1));

filldraw(circle((0,0),6), gray, black+linewidth(1));

filldraw(circle((0,0),4), white, black+linewidth(1));

So, the largest circle is filled light gray, then the middle circle (radius 6) is filled gray, which would cover the inner part? Wait, but when you draw a circle on top of another, the later one covers the previous. So the innermost circle (radius 4) is white, so that would be the area inside radius 4 is white, then the middle circle (radius 6) is gray, so the area between 4 and 6 is gray, and the outermost circle (radius 7) is light gray, so the area between 6 and 7 is light gray. Wait, but the Asymptote code draws the circles in order: first the largest (radius 7) is light gray, then the middle (radius 6) is gray, then the innermost (radius 4) is white. So the innermost circle is white, covering the center, then the middle circle (radius 6) is drawn on top, so the area between 4 and 6 is gray, and then the outermost circle (radius 7) is drawn on top, so the area between 6 and 7 is light gray. Therefore, the regions:

- X is the innermost white circle (radius 4).

- Y is the area between radius 4 and 6 (gray).

- Z is the area between radius 6 and 7 (light gray).

But the labels are placed at (2,0), (5,0), (6.5,0). So X is near the center, Y is between 4 and 6, and Z is between 6 and 7. So the problem says "three regions are labeled X, Y, or Z below." So the regions are:

- X: area of radius 4 (white)

- Y: area between 4 and 6 (gray)

- Z: area between 6 and 7 (light gray)

Therefore, to find the difference between the largest and smallest area among these three regions.

So first, let me compute the area of each region.

First, the area of X: that's the area of the circle with radius 4. So π*(4)^2 = 16π.

Then, the area of Y: that's the area between radius 4 and 6. So that's the area of the circle with radius 6 minus the area of the circle with radius 4. So π*(6)^2 - π*(4)^2 = 36π - 16π = 20π.

Then, the area of Z: that's the area between radius 6 and 7. So that's the area of the circle with radius 7 minus the area of the circle with radius 6. So π*(7)^2 - π*(6)^2 = 49π - 36π = 13π.

So the three areas are:

X: 16π

Y: 20π

Z: 13π

Therefore, the largest area is Y with 20π, and the smallest area is Z with 13π. Wait, but hold on, X is 16π, which is bigger than Z's 13π. Wait, so the regions are:

X: 16π

Y: 20π

Z: 13π

So the largest is Y (20π), the smallest is Z (13π). Therefore, the difference is 20π - 13π = 7π.

Wait, but hold on, is that correct? Let me check again.

Wait, the problem says "the difference between the area of the region with the greatest area and the area of the region with the smallest area." So if the regions are X, Y, Z, then the areas are:

X: inner circle (radius 4): area π*(4)^2 = 16π

Y: annulus between 4 and 6: area π*(6² - 4²) = π*(36 - 16) = 20π

Z: annulus between 6 and 7: area π*(7² - 6²) = π*(49 - 36) = 13π

So yes, the areas are 16π, 20π, 13π. Therefore, the largest is 20π (Y), the smallest is 13π (Z). The difference is 20π - 13π = 7π. Therefore, the answer is 7π.

But let me check again if I interpreted the regions correctly. The Asymptote code labels X at (2,0), which is on the x-axis, 2 units from the center. Since the inner circle has radius 4, so (2,0) is inside the inner circle. But the label is for region X. Wait, but if the inner circle is white, then maybe X is the white region? But then Y is labeled at (5,0), which is between 4 and 6, so that would be the gray region. Z is labeled at (6.5,0), which is between 6 and 7, so the light gray region. So yes, that seems correct. So the regions are:

X: inner circle (radius 4)

Y: between 4 and 6

Z: between 6 and 7

Therefore, their areas are as calculated. So the difference between the largest (Y: 20π) and smallest (Z: 13π) is 7π.

But wait, hold on. Wait, the problem says "three regions are labeled X, Y, or Z below." Maybe the regions are different? Maybe X is the area between 0 and 4, Y between 4 and 6, Z between 6 and 7. But maybe the problem is referring to different regions? Wait, but the Asymptote code shows that the innermost circle is white, so maybe X is the white region, Y is the gray region (between 4 and 6), and Z is the light gray region (between 6 and 7). So that's how the regions are divided. Therefore, the areas are as calculated.

Alternatively, maybe the regions are different? For example, maybe X is the area between 0 and 4, Y between 4 and 6, Z between 6 and 7. So that's the same as above. So the areas are 16π, 20π, 13π. So difference between largest and smallest is 20π - 13π = 7π.

But let me check once more. Maybe I made a mistake in the areas. Let me recalculate:

Area of X: π*(4)^2 = 16π. Correct.

Area of Y: π*(6)^2 - π*(4)^2 = 36π - 16π = 20π. Correct.

Area of Z: π*(7)^2 - π*(6)^2 = 49π - 36π = 13π. Correct.

So the difference between the largest (Y: 20π) and smallest (Z: 13π) is 7π. Therefore, the answer is 7π.

But let me check if the problem says "the difference between the area of the region with the greatest area and the area of the region with the smallest area." So if the regions are X, Y, Z, then the areas are 16π, 20π, 13π. So the greatest is 20π, the smallest is 13π. So difference is 7π.

Alternatively, maybe the regions are different? For example, maybe X is the area between 0 and 4, Y between 4 and 6, Z between 6 and 7. But that's the same as above.

Alternatively, maybe the regions are different? For example, maybe X is the area between 0 and 4, Y is the area between 0 and 6, and Z is the area between 0 and 7? But that would not make sense with the labels. The Asymptote code labels X at (2,0), which is inside the inner circle, so X is the inner circle. Y is at (5,0), which is between 4 and 6, so that's the middle annulus. Z is at (6.5,0), which is between 6 and 7, so the outer annulus. Therefore, the regions are as I thought.

Alternatively, maybe the problem is referring to the regions as follows: X is the area inside the smallest circle, Y is the area inside the middle circle but outside the smallest, and Z is the area inside the largest circle but outside the middle. Which is exactly what I considered.

Therefore, I think my calculation is correct. The difference is 7π.

But let me check once more. Maybe the problem is in the interpretation of the regions. For example, if the regions are not annuli but something else? But the Asymptote code draws three concentric circles, with the innermost white, middle gray, outer light gray. So the regions are:

- White: inside radius 4 (X)

- Gray: between 4 and 6 (Y)

- Light gray: between 6 and 7 (Z)

Therefore, the areas are as calculated.

Alternatively, maybe the problem is referring to the regions as the three different colored regions? So X is white, Y is gray, Z is light gray. Therefore, their areas are 16π, 20π, 13π.

Therefore, the answer is 20π - 13π = 7π.

Alternatively, maybe the problem is in the Asymptote code? Let me check the Asymptote code again:

filldraw(circle((0,0),7), lightgray, black+linewidth(1));

filldraw(circle((0,0),6), gray, black+linewidth(1));

filldraw(circle((0,0),4), white, black+linewidth(1));

So, the first command fills the largest circle (radius 7) with lightgray. Then, the second command fills the circle with radius 6 with gray. Since it's drawn on top, the area of radius 6 is gray, which would cover the inner part. Then, the third command fills the circle with radius 4 with white, which would cover the inner part again. Therefore, the regions:

- The innermost circle (radius 4) is white.

- The area between radius 4 and 6 is gray (since the middle circle is drawn on top of the largest circle, but the inner part is covered by white).

Wait, no. Wait, when you draw the largest circle (radius 7) as lightgray, then draw the middle circle (radius 6) as gray. So the middle circle (radius 6) is drawn on top of the lightgray circle. Therefore, the area inside radius 6 is gray, and the area between 6 and 7 is lightgray. Then, drawing the inner circle (radius 4) as white, which covers the area inside radius 4 with white. Therefore, the regions:

- Inside radius 4: white (X)

- Between 4 and 6: gray (Y)

- Between 6 and 7: lightgray (Z)

Therefore, the areas are as calculated.

Therefore, the answer is 7π.

But let me check once more. Maybe the problem is in the labels? The Asymptote code labels X at (2,0), which is inside the inner circle. Y at (5,0), which is between 4 and 6. Z at (6.5,0), which is between 6 and 7. Therefore, the regions are correctly labeled.

Therefore, I think my answer is correct. The difference is 7π.

But just to be thorough, let me check if there's another way to interpret the regions. For example, maybe X is the area between 0 and 4, Y between 0 and 6, Z between 0 and 7? But then the areas would be:

X: 16π

Y: 36π

Z: 49π

Then the difference would be 49π - 16π = 33π. But that seems unlikely because the labels are placed at (2,0), (5,0), (6.5,0). If X was between 0 and 4, Y between 0 and 6, Z between 0 and 7, then the labels would be placed in the middle of each region. For example, X would be labeled near the center, Y near the middle, Z near the outer edge. But the labels are at (2,0), (5,0), (6.5,0). So (2,0) is in the inner circle, (5,0) is between 4 and 6, (6.5,0) is between 6 and 7. Therefore, the regions are as I first thought.

Alternatively, maybe the regions are different? For example, maybe X is the area between 0 and 4, Y is the area between 4 and 7, and Z is something else? But the Asymptote code labels three regions, and the diagram shows three regions: white, gray, lightgray. So three regions. Therefore, the three regions are the three annular regions.

Therefore, I think my original calculation is correct. The answer is 7π.

But just to make sure, let me check the areas again:

- Area of X (radius 4): π*(4)^2 = 16π

- Area of Y (between 4 and 6): π*(6² - 4²) = π*(36 - 16) = 20π

- Area of Z (between 6 and 7): π*(7² - 6²) = π*(49 - 36) = 13π

So the areas are 16π, 20π, 13π. The largest is 20π, the smallest is 13π. Difference is 7π.

Yes, that seems correct. Therefore, the answer is 7π.

**Final Answer**
\boxed{7\pi}

To solve the problem, we analyze the three concentric circles with radii 4, 6, and 7. The diagram shows three regions labeled $ X $, $ Y $, and $ Z $, each corresponding to a distinct area between the circles. Based on the Asymptote code and the placement of the labels, we can deduce the following:

- **Region $ X $** is the innermost circle with radius 4.
- **Region $ Y $** is the annular region between radii 4 and 6.
- **Region $ Z $** is the annular region between radii 6 and 7.

---

### Step 1: Calculate the Area of Each Region

We use the formula for the area of a circle:
$$
A = \pi r^2
$$

- **Area of Region $ X $** (inner circle with radius 4):
$$
A_X = \pi (4)^2 = 16\pi
$$

- **Area of Region $ Y $** (annulus between radii 4 and 6):
$$
A_Y = \pi (6^2 - 4^2) = \pi (36 - 16) = 20\pi
$$

- **Area of Region $ Z $** (annulus between radii 6 and 7):
$$
A_Z = \pi (7^2 - 6^2) = \pi (49 - 36) = 13\pi
$$

---

### Step 2: Identify the Largest and Smallest Areas

From the calculated areas:

- $ A_X = 16\pi $
- $ A_Y = 20\pi $
- $ A_Z = 13\pi $

The **largest** area is $ A_Y = 20\pi $, and the **smallest** area is $ A_Z = 13\pi $.

---

### Step 3: Compute the Difference

$$
\text{Difference} = A_Y - A_Z = 20\pi - 13\pi = 7\pi
$$

---

### Final Answer

$$
\boxed{7\pi}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the difference between the area of the region with the greatest area and the area of the region with the smallest area among the three regions labeled X, Y, and Z in the diagram. The diagram shows three concentric circles with radii 4, 6, and 7.

First, let me visualize the diagram. There are three circles with the same center, right? The innermost circle has radius 4, then the middle one has radius 6, and the outermost one has radius 7. The regions X, Y, Z are probably the areas between these circles. The Asymptote code labels X at (2,0), which is probably the area between radius 0 and 4, Y at (5,0), which might be between 4 and 6, and Z at (6.5,0), which is between 6 and 7. Wait, but the Asymptote code fills the circles with different colors: the largest circle (radius 7) is light gray, the middle one (radius 6) is gray, and the innermost (radius 4) is white. So maybe the regions are:

- Region X: the white circle (radius 4), which is the innermost area.

- Region Y: the area between radius 4 and 6, which is the gray part minus the white part?

Wait, but the Asymptote code says:

filldraw(circle((0,0),7), lightgray, black+linewidth(1));

filldraw(circle((0,0),6), gray, black+linewidth(1));

filldraw(circle((0,0),4), white, black+linewidth(1));

So, the largest circle is filled light gray, then the middle circle (radius 6) is filled gray, which would cover the inner part? Wait, but when you draw a circle on top of another, the later one covers the previous. So the innermost circle (radius 4) is white, so that would be the area inside radius 4 is white, then the middle circle (radius 6) is gray, so the area between 4 and 6 is gray, and the outermost circle (radius 7) is light gray, so the area between 6 and 7 is light gray. Wait, but the Asymptote code draws the circles in order: first the largest (radius 7) is light gray, then the middle (radius 6) is gray, then the innermost (radius 4) is white. So the innermost circle is white, covering the center, then the middle circle (radius 6) is drawn on top, so the area between 4 and 6 is gray, and then the outermost circle (radius 7) is drawn on top, so the area between 6 and 7 is light gray. Therefore, the regions:

- X is the innermost white circle (radius 4).

- Y is the area between radius 4 and 6 (gray).

- Z is the area between radius 6 and 7 (light gray).

But the labels are placed at (2,0), (5,0), (6.5,0). So X is near the center, Y is between 4 and 6, and Z is between 6 and 7. So the problem says "three regions are labeled X, Y, or Z below." So the regions are:

- X: area of radius 4 (white)

- Y: area between 4 and 6 (gray)

- Z: area between 6 and 7 (light gray)

Therefore, to find the difference between the largest and smallest area among these three regions.

So first, let me compute the area of each region.

First, the area of X: that's the area of the circle with radius 4. So π*(4)^2 = 16π.

Then, the area of Y: that's the area between radius 4 and 6. So that's the area of the circle with radius 6 minus the area of the circle with radius 4. So π*(6)^2 - π*(4)^2 = 36π - 16π = 20π.

Then, the area of Z: that's the area between radius 6 and 7. So that's the area of the circle with radius 7 minus the area of the circle with radius 6. So π*(7)^2 - π*(6)^2 = 49π - 36π = 13π.

So the three areas are:

X: 16π

Y: 20π

Z: 13π

Therefore, the largest area is Y with 20π, and the smallest area is Z with 13π. Wait, but hold on, X is 16π, which is bigger than Z's 13π. Wait, so the regions are:

X: 16π

Y: 20π

Z: 13π

So the largest is Y (20π), the smallest is Z (13π). Therefore, the difference is 20π - 13π = 7π.

Wait, but hold on, is that correct? Let me check again.

Wait, the problem says "the difference between the area of the region with the greatest area and the area of the region with the smallest area." So if the regions are X, Y, Z, then the areas are:

X: inner circle (radius 4): area π*(4)^2 = 16π

Y: annulus between 4 and 6: area π*(6² - 4²) = π*(36 - 16) = 20π

Z: annulus between 6 and 7: area π*(7² - 6²) = π*(49 - 36) = 13π

So yes, the areas are 16π, 20π, 13π. Therefore, the largest is 20π (Y), the smallest is 13π (Z). The difference is 20π - 13π = 7π. Therefore, the answer is 7π.

But let me check again if I interpreted the regions correctly. The Asymptote code labels X at (2,0), which is on the x-axis, 2 units from the center. Since the inner circle has radius 4, so (2,0) is inside the inner circle. But the label is for region X. Wait, but if the inner circle is white, then maybe X is the white region? But then Y is labeled at (5,0), which is between 4 and 6, so that would be the gray region. Z is labeled at (6.5,0), which is between 6 and 7, so the light gray region. So yes, that seems correct. So the regions are:

X: inner circle (radius 4)

Y: between 4 and 6

Z: between 6 and 7

Therefore, their areas are as calculated. So the difference between the largest (Y: 20π) and smallest (Z: 13π) is 7π.

But wait, hold on. Wait, the problem says "three regions are labeled X, Y, or Z below." Maybe the regions are different? Maybe X is the area between 0 and 4, Y between 4 and 6, Z between 6 and 7. But maybe the problem is referring to different regions? Wait, but the Asymptote code shows that the innermost circle is white, so maybe X is the white region, Y is the gray region (between 4 and 6), and Z is the light gray region (between 6 and 7). So that's how the regions are divided. Therefore, the areas are as calculated.

Alternatively, maybe the regions are different? For example, maybe X is the area between 0 and 4, Y between 4 and 6, Z between 6 and 7. So that's the same as above. So the areas are 16π, 20π, 13π. So difference between largest and smallest is 20π - 13π = 7π.

But let me check once more. Maybe I made a mistake in the areas. Let me recalculate:

Area of X: π*(4)^2 = 16π. Correct.

Area of Y: π*(6)^2 - π*(4)^2 = 36π - 16π = 20π. Correct.

Area of Z: π*(7)^2 - π*(6)^2 = 49π - 36π = 13π. Correct.

So the difference between the largest (Y: 20π) and smallest (Z: 13π) is 7π. Therefore, the answer is 7π.

But let me check if the problem says "the difference between the area of the region with the greatest area and the area of the region with the smallest area." So if the regions are X, Y, Z, then the areas are 16π, 20π, 13π. So the greatest is 20π, the smallest is 13π. So difference is 7π.

Alternatively, maybe the regions are different? For example, maybe X is the area between 0 and 4, Y between 4 and 6, Z between 6 and 7. But that's the same as above.

Alternatively, maybe the regions are different? For example, maybe X is the area between 0 and 4, Y is the area between 0 and 6, and Z is the area between 0 and 7? But that would not make sense with the labels. The Asymptote code labels X at (2,0), which is inside the inner circle, so X is the inner circle. Y is at (5,0), which is between 4 and 6, so that's the middle annulus. Z is at (6.5,0), which is between 6 and 7, so the outer annulus. Therefore, the regions are as I thought.

Alternatively, maybe the problem is referring to the regions as follows: X is the area inside the smallest circle, Y is the area inside the middle circle but outside the smallest, and Z is the area inside the largest circle but outside the middle. Which is exactly what I considered.

Therefore, I think my calculation is correct. The difference is 7π.

But let me check once more. Maybe the problem is in the interpretation of the regions. For example, if the regions are not annuli but something else? But the Asymptote code draws three concentric circles, with the innermost white, middle gray, outer light gray. So the regions are:

- White: inside radius 4 (X)

- Gray: between 4 and 6 (Y)

- Light gray: between 6 and 7 (Z)

Therefore, the areas are as calculated.

Alternatively, maybe the problem is referring to the regions as the three different colored regions? So X is white, Y is gray, Z is light gray. Therefore, their areas are 16π, 20π, 13π.

Therefore, the answer is 20π - 13π = 7π.

Alternatively, maybe the problem is in the Asymptote code? Let me check the Asymptote code again:

filldraw(circle((0,0),7), lightgray, black+linewidth(1));

filldraw(circle((0,0),6), gray, black+linewidth(1));

filldraw(circle((0,0),4), white, black+linewidth(1));

So, the first command fills the largest circle (radius 7) with lightgray. Then, the second command fills the circle with radius 6 with gray. Since it's drawn on top, the area of radius 6 is gray, which would cover the inner part. Then, the third command fills the circle with radius 4 with white, which would cover the inner part again. Therefore, the regions:

- The innermost circle (radius 4) is white.

- The area between radius 4 and 6 is gray (since the middle circle is drawn on top of the largest circle, but the inner part is covered by white).

Wait, no. Wait, when you draw the largest circle (radius 7) as lightgray, then draw the middle circle (radius 6) as gray. So the middle circle (radius 6) is drawn on top of the lightgray circle. Therefore, the area inside radius 6 is gray, and the area between 6 and 7 is lightgray. Then, drawing the inner circle (radius 4) as white, which covers the area inside radius 4 with white. Therefore, the regions:

- Inside radius 4: white (X)

- Between 4 and 6: gray (Y)

- Between 6 and 7: lightgray (Z)

Therefore, the areas are as calculated.

Therefore, the answer is 7π.

But let me check once more. Maybe the problem is in the labels? The Asymptote code labels X at (2,0), which is inside the inner circle. Y at (5,0), which is between 4 and 6. Z at (6.5,0), which is between 6 and 7. Therefore, the regions are correctly labeled.

Therefore, I think my answer is correct. The difference is 7π.

But just to be thorough, let me check if there's another way to interpret the regions. For example, maybe X is the area between 0 and 4, Y between 0 and 6, Z between 0 and 7? But then the areas would be:

X: 16π

Y: 36π

Z: 49π

Then the difference would be 49π - 16π = 33π. But that seems unlikely because the labels are placed at (2,0), (5,0), (6.5,0). If X was between 0 and 4, Y between 0 and 6, Z between 0 and 7, then the labels would be placed in the middle of each region. For example, X would be labeled near the center, Y near the middle, Z near the outer edge. But the labels are at (2,0), (5,0), (6.5,0). So (2,0) is in the inner circle, (5,0) is between 4 and 6, (6.5,0) is between 6 and 7. Therefore, the regions are as I first thought.

Alternatively, maybe the regions are different? For example, maybe X is the area between 0 and 4, Y is the area between 4 and 7, and Z is something else? But the Asymptote code labels three regions, and the diagram shows three regions: white, gray, lightgray. So three regions. Therefore, the three regions are the three annular regions.

Therefore, I think my original calculation is correct. The answer is 7π.

But just to make sure, let me check the areas again:

- Area of X (radius 4): π*(4)^2 = 16π

- Area of Y (between 4 and 6): π*(6² - 4²) = π*(36 - 16) = 20π

- Area of Z (between 6 and 7): π*(7² - 6²) = π*(49 - 36) = 13π

So the areas are 16π, 20π, 13π. The largest is 20π, the smallest is 13π. Difference is 7π.

Yes, that seems correct. Therefore, the answer is 7π.

**Final Answer**
\boxed{7\pi}

To solve the problem, we analyze the three concentric circles with radii 4, 6, and 7. The diagram shows three regions labeled $ X $, $ Y $, and $ Z $, each corresponding to a distinct area between the circles. Based on the Asymptote code and the placement of the labels, we can deduce the following:

- **Region $ X $** is the innermost circle with radius 4.
- **Region $ Y $** is the annular region between radii 4 and 6.
- **Region $ Z $** is the annular region between radii 6 and 7.

---

### Step 1: Calculate the Area of Each Region

We use the formula for the area of a circle:
$$
A = \pi r^2
$$

- **Area of Region $ X $** (inner circle with radius 4):
$$
A_X = \pi (4)^2 = 16\pi
$$

- **Area of Region $ Y $** (annulus between radii 4 and 6):
$$
A_Y = \pi (6^2 - 4^2) = \pi (36 - 16) = 20\pi
$$

- **Area of Region $ Z $** (annulus between radii 6 and 7):
$$
A_Z = \pi (7^2 - 6^2) = \pi (49 - 36) = 13\pi
$$

---

### Step 2: Identify the Largest and Smallest Areas

From the calculated areas:

- $ A_X = 16\pi $
- $ A_Y = 20\pi $
- $ A_Z = 13\pi $

The **largest** area is $ A_Y = 20\pi $, and the **smallest** area is $ A_Z = 13\pi $.

---

### Step 3: Compute the Difference

$$
\text{Difference} = A_Y - A_Z = 20\pi - 13\pi = 7\pi
$$

---

### Final Answer

$$
\boxed{7\pi}
$$

Solution:
The area of the inside circle (region $X$) is $\pi\cdot 4^2=16\pi.$

Using a similar technique, the area of the middle ring (region $Y$) is $$\pi\cdot 6^2-\pi\cdot 4^2=36\pi-16\pi = 20\pi.$$ Also, the area of the outer ring (region $Z$) is $$\pi\cdot 7^2-\pi\cdot 6^2=49\pi - 36\pi = 13\pi.$$ Therefore, region $Y$ has the largest area and region $Z$ has the smallest area. The difference in their areas is $20\pi-13\pi = \boxed{7\pi}.$


Explanation:
Okay, so I need to solve the equation:

$$
\sin \left( \tan^{-1} (x) + \cot^{-1} \left( \frac{1}{x} \right) \right) = \frac{1}{3}.
$$

Hmm, let me start by recalling some trigonometric identities and inverse trigonometric functions properties. Maybe I can simplify the expression inside the sine function first. Let me denote:

Let’s let θ = tan⁻¹(x). Then tan θ = x. Also, cot⁻¹(1/x) is another term. I remember that cot⁻¹(y) is equal to tan⁻¹(1/y), right? Wait, is that true?

Wait, let me check. The cotangent is the reciprocal of tangent, so cot(α) = 1/tan(α). Therefore, if cot⁻¹(y) is the angle whose cotangent is y, then that angle is equal to tan⁻¹(1/y). So yes, cot⁻¹(y) = tan⁻¹(1/y). Therefore, cot⁻¹(1/x) = tan⁻¹(x). Wait, that would mean that cot⁻¹(1/x) is tan⁻¹(x). But if θ = tan⁻¹(x), then cot⁻¹(1/x) is also θ? Let me verify with an example. Suppose x = 1. Then tan⁻¹(1) is π/4, and cot⁻¹(1/1) = cot⁻¹(1) which is also π/4. So yes, that seems to hold. So in general, cot⁻¹(1/x) = tan⁻¹(x). Therefore, the argument inside the sine function is tan⁻¹(x) + tan⁻¹(x) = 2 tan⁻¹(x). Therefore, the equation simplifies to:

sin(2 tan⁻¹(x)) = 1/3.

Wait, is that correct? Let me check again. If cot⁻¹(1/x) = tan⁻¹(x), then adding tan⁻¹(x) + cot⁻¹(1/x) is tan⁻¹(x) + tan⁻¹(x) = 2 tan⁻¹(x). Therefore, the equation becomes sin(2 tan⁻¹(x)) = 1/3. That seems right. So maybe I can use the double angle formula for sine here. Let me recall that sin(2α) = 2 sin α cos α. So if I let α = tan⁻¹(x), then sin(2α) = 2 sin α cos α. But since α = tan⁻¹(x), then we can think of a right triangle where the opposite side is x and adjacent side is 1, so hypotenuse is √(1 + x²). Therefore, sin α = x / √(1 + x²) and cos α = 1 / √(1 + x²). Therefore, sin(2α) = 2*(x / √(1 + x²))*(1 / √(1 + x²)) = 2x / (1 + x²). Therefore, the equation becomes:

2x / (1 + x²) = 1/3.

So now, solving 2x / (1 + x²) = 1/3. Let me cross multiply:

2x * 3 = 1 * (1 + x²)

Which is 6x = 1 + x²

Rearranging terms:

x² - 6x + 1 = 0

Wait, hold on. Let me check that again. If 6x = 1 + x², then bringing all terms to one side:

x² - 6x + 1 = 0? Wait, no. If 6x = 1 + x², then subtract 6x from both sides:

0 = x² - 6x + 1. Yes, that's correct. So quadratic equation x² - 6x + 1 = 0. Let me solve this quadratic equation. Using quadratic formula:

x = [6 ± √(36 - 4*1*1)] / 2 = [6 ± √(32)] / 2 = [6 ± 4√2]/2 = 3 ± 2√2.

So the solutions are x = 3 + 2√2 and x = 3 - 2√2. Wait, but I need to check if these solutions are valid in the original equation. Because sometimes when dealing with inverse trigonometric functions, there might be restrictions on the domain or some extraneous solutions introduced.

But let me think. The original equation involves tan⁻¹(x) and cot⁻¹(1/x). The domain of tan⁻¹(x) is all real numbers, and cot⁻¹(1/x) requires that 1/x is defined, so x ≠ 0. Therefore, the domain of the original equation is x ≠ 0. So our solutions 3 + 2√2 and 3 - 2√2 are both non-zero, so they are in the domain. But maybe there's something else?

Wait, but let me check if there are any restrictions on the range of the inverse functions. For example, tan⁻¹(x) gives values between -π/2 and π/2, and cot⁻¹(1/x) gives values between 0 and π. However, when we added them together, we need to make sure that the sum is within the domain where sine is defined, which it always is. But maybe there's a problem with the identity I used earlier?

Wait, earlier I said that cot⁻¹(1/x) = tan⁻¹(x). Let me check if that's always true. Let me consider x positive first. If x is positive, then tan⁻¹(x) is between 0 and π/2, and cot⁻¹(1/x) is also between 0 and π/2. Since cot⁻¹(y) = tan⁻¹(1/y) for y > 0, so if y = 1/x, then cot⁻¹(1/x) = tan⁻¹(x). So that's correct for x > 0. What if x is negative?

Suppose x is negative. Let me take x = -a where a > 0. Then tan⁻¹(x) = tan⁻¹(-a) = -tan⁻¹(a). On the other hand, cot⁻¹(1/x) = cot⁻¹(-1/a). But cot⁻¹(y) for negative y. Wait, the range of cot⁻¹(y) is typically (0, π). So cot⁻¹(-1/a) would be π - cot⁻¹(1/a). Because cotangent is negative in the second quadrant. Let me recall that cot⁻¹(y) is defined as the angle in (0, π) whose cotangent is y. So if y is negative, then cot⁻¹(y) is in (π/2, π). So for example, cot⁻¹(-1/a) = π - cot⁻¹(1/a). But cot⁻¹(1/a) is tan⁻¹(a). Therefore, cot⁻¹(-1/a) = π - tan⁻¹(a). Therefore, cot⁻¹(1/x) when x is negative is π - tan⁻¹(a) = π - tan⁻¹(-x). Therefore, in that case, tan⁻¹(x) + cot⁻¹(1/x) = tan⁻¹(-a) + π - tan⁻¹(a) = -tan⁻¹(a) + π - tan⁻¹(a) = π - 2 tan⁻¹(a). But if we consider the original expression, when x is negative, tan⁻¹(x) + cot⁻¹(1/x) is π - 2 tan⁻¹(-x). Wait, but in our previous step, we assumed that cot⁻¹(1/x) = tan⁻¹(x), which is only true when x is positive. For negative x, that identity doesn't hold. Therefore, my earlier conclusion that tan⁻¹(x) + cot⁻¹(1/x) = 2 tan⁻¹(x) is only valid when x is positive. If x is negative, then the sum is π - 2 tan⁻¹(-x). Therefore, we need to check if x is positive or negative.

Therefore, perhaps my initial simplification was only valid for x > 0. Therefore, maybe I need to consider two cases: x > 0 and x < 0.

So let me re-examine the original problem. Let me first check if x is positive or negative.

Case 1: x > 0.

In this case, as we saw, cot⁻¹(1/x) = tan⁻¹(x). Therefore, tan⁻¹(x) + cot⁻¹(1/x) = 2 tan⁻¹(x). Therefore, sin(2 tan⁻¹(x)) = 1/3. Which led us to the equation 2x/(1 + x²) = 1/3, which gave solutions x = 3 ± 2√2. However, since x is positive, we need to check which of these solutions are positive. 3 + 2√2 is definitely positive. 3 - 2√2 is approximately 3 - 2.828 ≈ 0.172, which is also positive. Therefore, both solutions are positive. Therefore, both are valid in this case.

Case 2: x < 0.

In this case, as we saw, tan⁻¹(x) + cot⁻¹(1/x) = π - 2 tan⁻¹(-x). Let me denote y = -x, so y > 0. Then tan⁻¹(x) + cot⁻¹(1/x) = π - 2 tan⁻¹(y). Therefore, the original equation becomes:

sin(π - 2 tan⁻¹(y)) = 1/3.

But sin(π - θ) = sin θ. Therefore, sin(π - 2 tan⁻¹(y)) = sin(2 tan⁻¹(y)) = 1/3. Therefore, this reduces to the same equation as in Case 1: sin(2 tan⁻¹(y)) = 1/3. Which again leads to 2y/(1 + y²) = 1/3. Therefore, solving for y gives y = 3 ± 2√2. But since y = -x and x < 0, then y > 0. Therefore, y can be 3 + 2√2 or 3 - 2√2. Therefore, x = -y, so x = -3 - 2√2 or x = -3 + 2√2. However, we need to check if these solutions are valid. Let me check if they satisfy the original equation.

But wait, let me think again. Since we have sin(π - 2 tan⁻¹(y)) = sin(2 tan⁻¹(y)) = 1/3. Therefore, the equation reduces to the same equation as in Case 1. Therefore, solving for y gives the same solutions. Therefore, in Case 2, we have x = -y, so x = - (3 + 2√2) and x = - (3 - 2√2). However, we need to check if these solutions are valid. Let me check if they satisfy the original equation.

But let me check if there's any problem with the steps. For example, when we have sin(π - θ) = sin θ, which is true. Therefore, if we have sin(π - 2 tan⁻¹(y)) = 1/3, then sin(2 tan⁻¹(y)) = 1/3. Therefore, solving for y gives the same equation as before. Therefore, the solutions for y are 3 ± 2√2, so x = - (3 ± 2√2). Therefore, in total, we have four solutions? Wait, but in Case 1, x > 0 gives two solutions, and in Case 2, x < 0 gives two solutions. But let me check if all four solutions are valid.

Wait, but let me check if when x is negative, the original equation is valid. Let me take x = - (3 - 2√2). Let me compute 3 - 2√2. Since √2 ≈ 1.414, so 2√2 ≈ 2.828. Therefore, 3 - 2.828 ≈ 0.172. Therefore, x ≈ -0.172. Let me check if this satisfies the original equation.

Alternatively, maybe there's a mistake here. Let me think again. When we have x negative, we have tan⁻¹(x) + cot⁻¹(1/x) = π - 2 tan⁻¹(-x). Then sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x))? Wait, no. Wait, sin(π - θ) = sin θ. Therefore, sin(π - 2 tan⁻¹(y)) = sin(2 tan⁻¹(y)). Therefore, the equation becomes sin(2 tan⁻¹(y)) = 1/3. Therefore, solving for y gives the same equation as before. Therefore, the solutions for y are 3 ± 2√2, so x = -y. Therefore, x = -3 - 2√2 and x = -3 + 2√2. However, we need to check if these solutions are valid. Let me check if they satisfy the original equation.

But let me check if there's a problem with the range of the inverse functions. For example, when x is negative, tan⁻¹(x) is between -π/2 and 0, and cot⁻¹(1/x) is between π/2 and π. Therefore, their sum is between 0 and π. Therefore, sin of that sum is between 0 and 1. Since 1/3 is positive, that's okay. Therefore, the solutions should be valid. Therefore, in total, there are four solutions? Wait, but when I solved the equation, I had x² - 6x + 1 = 0, which gave two solutions. But if we consider both positive and negative x, then we have four solutions? Wait, no. Wait, the equation 2x/(1 + x²) = 1/3 led to x² - 6x + 1 = 0, which has two solutions. However, when we considered x negative, we found that the equation reduces to the same equation for y, which is positive. Therefore, the solutions for x negative are - (3 + 2√2) and - (3 - 2√2). Therefore, in total, there are four solutions? Wait, but the equation 2x/(1 + x²) = 1/3 is a quadratic equation, which can have at most two real solutions. Therefore, how come we have four solutions? There must be a mistake here.

Wait, no. Wait, when I considered x negative, I transformed the original equation into the same equation as for x positive. Therefore, solving 2x/(1 + x²) = 1/3 gives two solutions, but when x is negative, we have to check if they satisfy the original equation. Wait, but if we have x negative, then the equation sin(π - 2 tan⁻¹(y)) = 1/3 becomes sin(2 tan⁻¹(y)) = 1/3, which is the same equation as before. Therefore, solving for y gives the same solutions. Therefore, the solutions for x negative are - (3 + 2√2) and - (3 - 2√2). However, when we plug these into the equation 2x/(1 + x²) = 1/3, do they satisfy?

Let me check. Let me take x = - (3 + 2√2). Let me compute 2x/(1 + x²). Let me compute x²:

x = - (3 + 2√2). Then x² = (3 + 2√2)^2 = 9 + 12√2 + 8 = 17 + 12√2.

Then 1 + x² = 18 + 12√2.

Then 2x = -2*(3 + 2√2) = -6 - 4√2.

Therefore, 2x/(1 + x²) = (-6 - 4√2)/(18 + 12√2). Let me rationalize the denominator or simplify:

Factor numerator and denominator:

Numerator: -2*(3 + 2√2)

Denominator: 6*(3 + 2√2)

Therefore, the ratio is (-2/6) = -1/3. But the original equation is sin(...) = 1/3. However, we have 2x/(1 + x²) = -1/3. But earlier, we had that sin(2 tan⁻¹(x)) = 2x/(1 + x²). However, when x is negative, we had that the original equation reduces to sin(π - 2 tan⁻¹(y)) = 1/3, which is equal to sin(2 tan⁻¹(y)) = 1/3. Therefore, in this case, the value of 2x/(1 + x²) is -1/3, but the sine of the angle is 1/3. Therefore, there is a contradiction here. Therefore, my previous reasoning was flawed.

Wait, so where is the mistake? Let me retrace. When x is negative, we have:

Original expression inside sine: tan⁻¹(x) + cot⁻¹(1/x) = π - 2 tan⁻¹(-x). Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x))? Wait, no. Wait, sin(π - θ) = sin θ. Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x)). Wait, but tan⁻¹(-x) is equal to - tan⁻¹(x). Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x)) = sin(-2 tan⁻¹(x)) = - sin(2 tan⁻¹(x)). Therefore, the original equation becomes - sin(2 tan⁻¹(x)) = 1/3. Therefore, sin(2 tan⁻¹(x)) = -1/3. Therefore, in this case, the equation is not 2x/(1 + x²) = 1/3, but rather 2x/(1 + x²) = -1/3. Therefore, my mistake was in the previous step where I thought that sin(π - 2 tan⁻¹(y)) = sin(2 tan⁻¹(y)), but actually, it's sin(π - θ) = sin θ, but θ here is 2 tan⁻¹(y). Wait, no. Wait, if θ = 2 tan⁻¹(y), then sin(π - θ) = sin θ. Therefore, sin(π - θ) = sin θ. Therefore, if the original equation is sin(π - θ) = 1/3, then sin θ = 1/3. Therefore, θ = arcsin(1/3) or π - arcsin(1/3). Therefore, in this case, θ = 2 tan⁻¹(y). Therefore, 2 tan⁻¹(y) = arcsin(1/3) or π - arcsin(1/3). Therefore, solving for y would give different solutions. Therefore, my previous conclusion that it reduces to the same equation was wrong. Therefore, I need to correct this.

Therefore, going back. For x < 0, we have:

Original equation: sin(π - 2 tan⁻¹(-x)) = 1/3. Let me denote θ = 2 tan⁻¹(-x). Then sin(π - θ) = sin θ = 1/3. Therefore, sin θ = 1/3. Therefore, θ = arcsin(1/3) or π - arcsin(1/3). Therefore, 2 tan⁻¹(-x) = arcsin(1/3) or π - arcsin(1/3). Therefore, tan⁻¹(-x) = (1/2) arcsin(1/3) or (π - arcsin(1/3))/2. Therefore, -x = tan[(1/2) arcsin(1/3)] or tan[(π - arcsin(1/3))/2]. Therefore, solving for x would give different expressions. However, this seems complicated. Alternatively, maybe I can relate this back to the equation 2x/(1 + x²) = -1/3. Because earlier, for x positive, we had sin(2 tan⁻¹(x)) = 2x/(1 + x²). For x negative, we have sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x)) = - sin(2 tan⁻¹(x)) = - [2x/(1 + x²)]? Wait, let me check.

If x is negative, then tan⁻¹(x) = - tan⁻¹(-x). Let me denote α = tan⁻¹(-x), which is positive because -x is positive. Then tan α = -x. Therefore, sin(2 tan⁻¹(x)) = sin(-2α) = - sin(2α). But 2α = 2 tan⁻¹(-x). Therefore, sin(π - 2α) = sin(2α) = 1/3. Therefore, sin(2α) = 1/3. Therefore, sin(2α) = 1/3. But 2α = 2 tan⁻¹(-x). Therefore, sin(2 tan⁻¹(-x)) = 1/3. But sin(2 tan⁻¹(-x)) = 2*(-x)/(1 + (-x)^2) = -2x/(1 + x²). Therefore, -2x/(1 + x²) = 1/3. Therefore, solving -2x/(1 + x²) = 1/3. Which leads to -2x = (1 + x²)/3 => -6x = 1 + x² => x² + 6x + 1 = 0. Therefore, solving x² + 6x + 1 = 0. Therefore, x = [-6 ± √(36 - 4)] / 2 = [-6 ± √32]/2 = [-6 ± 4√2]/2 = -3 ± 2√2. Therefore, x = -3 + 2√2 or x = -3 - 2√2. Therefore, these are the solutions for x < 0. Therefore, in total, the solutions are:

From x > 0: 3 + 2√2 and 3 - 2√2.

From x < 0: -3 + 2√2 and -3 - 2√2.

But wait, let me check if these solutions are valid. Let me check for x = -3 + 2√2. Let me compute approximate values. √2 ≈ 1.414, so 2√2 ≈ 2.828. Therefore, -3 + 2.828 ≈ -0.172. Similarly, -3 - 2√2 ≈ -5.828. Let me check if these satisfy the original equation.

First, take x = -3 + 2√2 ≈ -0.172. Let me compute tan⁻¹(x) + cot⁻¹(1/x). Let me compute tan⁻¹(x). Since x is negative, tan⁻¹(x) is negative. Let me compute cot⁻¹(1/x). Since x is negative, 1/x is negative. cot⁻¹(1/x) is in (π/2, π). Let me compute tan⁻¹(x) + cot⁻¹(1/x). Let me denote θ = tan⁻¹(x). Then cot⁻¹(1/x) = π - tan⁻¹(-1/x) = π - tan⁻¹(1/|x|). Wait, but maybe it's easier to compute numerically.

Alternatively, since we derived that for x negative, the equation reduces to -2x/(1 + x²) = 1/3. Let me check for x = -3 + 2√2. Let me compute -2x/(1 + x²):

First, compute x = -3 + 2√2. Let me compute x²:

x = -3 + 2√2. Then x² = (-3)^2 + (2√2)^2 - 2*3*2√2 = 9 + 8 - 12√2 = 17 - 12√2.

Then 1 + x² = 1 + 17 - 12√2 = 18 - 12√2.

Then -2x = -2*(-3 + 2√2) = 6 - 4√2.

Therefore, -2x/(1 + x²) = (6 - 4√2)/(18 - 12√2). Let me rationalize the denominator:

Multiply numerator and denominator by (18 + 12√2):

Numerator: (6 - 4√2)(18 + 12√2)

Denominator: (18 - 12√2)(18 + 12√2) = 18² - (12√2)² = 324 - 288 = 36.

Compute numerator:

First, expand (6)(18) + 6*(12√2) - 4√2*(18) - 4√2*(12√2)

= 108 + 72√2 - 72√2 - 48*(√2)^2

Simplify:

108 + (72√2 - 72√2) - 48*2

= 108 - 96 = 12.

Therefore, numerator is 12, denominator is 36. Therefore, 12/36 = 1/3. Therefore, -2x/(1 + x²) = 1/3. Therefore, this satisfies the equation. Therefore, x = -3 + 2√2 is a valid solution. Similarly, for x = -3 - 2√2, let me check:

x = -3 - 2√2. Compute -2x/(1 + x²):

First, x = -3 - 2√2. Then x² = (-3 - 2√2)^2 = 9 + 12√2 + 8 = 17 + 12√2.

Then 1 + x² = 18 + 12√2.

-2x = -2*(-3 - 2√2) = 6 + 4√2.

Therefore, -2x/(1 + x²) = (6 + 4√2)/(18 + 12√2). Again, rationalize denominator by multiplying numerator and denominator by (18 - 12√2):

Numerator: (6 + 4√2)(18 - 12√2)

Denominator: same as before, 36.

Compute numerator:

6*18 + 6*(-12√2) + 4√2*18 + 4√2*(-12√2)

= 108 - 72√2 + 72√2 - 48*(√2)^2

= 108 - 0 - 48*2 = 108 - 96 = 12.

Therefore, 12/36 = 1/3. Therefore, this also satisfies the equation. Therefore, both x = -3 + 2√2 and x = -3 - 2√2 are valid solutions.

Therefore, in total, the four solutions are:

x = 3 + 2√2,

x = 3 - 2√2,

x = -3 + 2√2,

x = -3 - 2√2.

But wait, let me check if these are all distinct. Let me compute approximate values:

3 + 2√2 ≈ 3 + 2.828 ≈ 5.828,

3 - 2√2 ≈ 3 - 2.828 ≈ 0.172,

-3 + 2√2 ≈ -3 + 2.828 ≈ -0.172,

-3 - 2√2 ≈ -3 - 2.828 ≈ -5.828.

So all four solutions are distinct. Therefore, the original equation has four solutions. However, let me check if there are any restrictions. For example, when x = 0, the original equation is undefined, but our solutions don't include 0. Also, for all other x, the functions are defined. Therefore, all four solutions are valid.

But wait, let me check with the original equation. Let me take x = 3 - 2√2 ≈ 0.172. Let me compute tan⁻¹(x) + cot⁻¹(1/x). Let me compute tan⁻¹(x) ≈ tan⁻¹(0.172) ≈ 0.17 radians (approx). Then 1/x ≈ 5.828, so cot⁻¹(5.828) ≈ tan⁻¹(1/5.828) ≈ tan⁻¹(0.172) ≈ 0.17 radians. Therefore, adding them together gives approx 0.34 radians. Then sin(0.34) ≈ 0.333, which is approx 1/3. Therefore, that works. Similarly, for x = -0.172, tan⁻¹(x) ≈ -0.17 radians, and cot⁻¹(1/x) = cot⁻¹(-5.828) which is π - tan⁻¹(5.828) ≈ π - 1.42 radians ≈ 1.72 radians. Adding tan⁻¹(x) + cot⁻¹(1/x) ≈ -0.17 + 1.72 ≈ 1.55 radians. Then sin(1.55) ≈ sin(π - 1.55) ≈ sin(1.59) ≈ 0.999? Wait, no. Wait, sin(1.55) is approx sin(π - 1.55) = sin(1.59) ≈ 0.999? Wait, no. Wait, π is approx 3.14, so π - 1.55 ≈ 1.59 radians. Wait, sin(1.55) is approx sin(π - 1.55) = sin(1.59). Wait, but sin(1.55) ≈ sin(88.8 degrees) ≈ 0.999. Wait, but 1.55 radians is approx 88.8 degrees? Wait, π/2 is approx 1.57 radians, so 1.55 radians is slightly less than π/2, so sin(1.55) ≈ sin(π/2 - 0.02) ≈ cos(0.02) ≈ 0.9998. But according to our equation, it should be 1/3 ≈ 0.333. Wait, this seems contradictory. What's wrong here?

Wait, there must be a miscalculation here. Let me check again. If x = -0.172, then tan⁻¹(x) ≈ -0.17 radians. Then cot⁻¹(1/x) = cot⁻¹(-5.828). Since cot⁻¹(y) for negative y is π - cot⁻¹(|y|). So cot⁻¹(-5.828) = π - cot⁻¹(5.828). But cot⁻¹(5.828) is tan⁻¹(1/5.828) ≈ tan⁻¹(0.172) ≈ 0.17 radians. Therefore, cot⁻¹(-5.828) ≈ π - 0.17 ≈ 3.14 - 0.17 ≈ 2.97 radians. Therefore, tan⁻¹(x) + cot⁻¹(1/x) ≈ -0.17 + 2.97 ≈ 2.8 radians. Then sin(2.8 radians). Let me compute sin(2.8). 2.8 radians is approx 160 degrees (since π radians is 180 degrees, so 2.8 radians ≈ 160 degrees). sin(160 degrees) = sin(20 degrees) ≈ 0.342, which is approx 1/3 (≈ 0.333). Therefore, that works. Therefore, sin(2.8) ≈ 0.342 ≈ 1/3. Therefore, that's correct. Therefore, the solution x = -0.172 is valid. Similarly, for x = -5.828, let me check. tan⁻¹(x) ≈ tan⁻¹(-5.828) ≈ -1.42 radians. cot⁻¹(1/x) = cot⁻¹(-1/5.828) = cot⁻¹(-0.172) = π - cot⁻¹(0.172) ≈ π - 0.17 ≈ 2.97 radians. Therefore, adding tan⁻¹(x) + cot⁻¹(1/x) ≈ -1.42 + 2.97 ≈ 1.55 radians. Then sin(1.55) ≈ sin(π - 1.55) = sin(1.59) ≈ 0.999? Wait, no. Wait, sin(1.55) is approx sin(88.8 degrees) ≈ 0.999, but according to our equation, it should be 1/3. Wait, this is conflicting. Wait, but earlier we had that for x negative, the equation reduces to -2x/(1 + x²) = 1/3. Let me check for x = -3 - 2√2 ≈ -5.828. Let me compute -2x/(1 + x²). As we did before, it equals 1/3. Therefore, according to our previous derivation, sin(π - 2 tan⁻¹(-x)) = 1/3. Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x)) = 1/3. Wait, but earlier I thought that sin(π - θ) = sin θ, so sin(π - θ) = sin θ. Therefore, if θ = 2 tan⁻¹(-x), then sin(π - θ) = sin θ = 1/3. Therefore, sin(θ) = 1/3. Therefore, θ = arcsin(1/3) or π - arcsin(1/3). Therefore, 2 tan⁻¹(-x) = arcsin(1/3) or π - arcsin(1/3). Therefore, tan⁻¹(-x) = (1/2) arcsin(1/3) or (π - arcsin(1/3))/2. Therefore, -x = tan[(1/2) arcsin(1/3)] or tan[(π - arcsin(1/3))/2]. Therefore, solving for x gives us the solutions we found. Therefore, even though when I compute numerically, I get sin(1.55) ≈ 0.999, which conflicts with 1/3, there must be an error in my numerical estimation. Wait, let me compute 1.55 radians. Let me convert 1.55 radians to degrees: 1.55 * (180/π) ≈ 1.55 * 57.3 ≈ 88.8 degrees. Wait, sin(88.8 degrees) is indeed approximately 0.999. But according to our equation, it should be 1/3. Therefore, there is a contradiction here. What is wrong?

Wait, let me check again. For x = -3 - 2√2 ≈ -5.828. Let me compute tan⁻¹(x) + cot⁻¹(1/x). Let me compute tan⁻¹(x) ≈ tan⁻¹(-5.828) ≈ -1.42 radians. Then cot⁻¹(1/x) = cot⁻¹(-1/5.828) = cot⁻¹(-0.172). As before, cot⁻¹(-0.172) = π - cot⁻¹(0.172) ≈ π - 0.17 ≈ 2.97 radians. Therefore, adding tan⁻¹(x) + cot⁻¹(1/x) ≈ -1.42 + 2.97 ≈ 1.55 radians. Then sin(1.55) ≈ 0.999. But according to the equation, it should be 1/3. Therefore, something is wrong here. But earlier, we had that -2x/(1 + x²) = 1/3. Let me check that for x = -3 - 2√2. As we did before, it gives 1/3. Therefore, according to our previous derivation, sin(π - 2 tan⁻¹(-x)) = 1/3. But π - 2 tan⁻¹(-x) is equal to π - 2 tan⁻¹(-x). Let me compute tan⁻¹(-x) where x = -3 - 2√2. So -x = 3 + 2√2. Therefore, tan⁻¹(-x) = tan⁻¹(3 + 2√2). Let me compute tan⁻¹(3 + 2√2). Let me note that tan(π/8) = √2 - 1 ≈ 0.414, tan(3π/8) = √2 + 1 ≈ 2.414. Wait, 3 + 2√2 ≈ 3 + 2.828 ≈ 5.828. Let me compute tan(π/2 - something). Wait, tan(π/2 - θ) = cot θ. Let me think. Alternatively, note that tan(π/4 + θ) = (1 + tan θ)/(1 - tan θ). Maybe there's an identity here. Alternatively, let me compute tan(π/2 - arctan(3 + 2√2))? Not sure. Alternatively, let me compute tan(π/2 - α) = cot α. If I can find α such that tan α = 3 + 2√2. Let me compute tan(π/8) = √2 - 1 ≈ 0.414, tan(3π/8) = √2 + 1 ≈ 2.414. tan(5π/8) = tan(π - 3π/8) = -tan(3π/8) ≈ -2.414. tan(7π/8) = tan(π - π/8) = -tan(π/8) ≈ -0.414. So tan(π/2 - α) = cot α. If I can find α such that tan α = 3 + 2√2. Let me compute tan(π/2 - α) = cot α = 3 + 2√2. Therefore, π/2 - α = arccot(3 + 2√2). But I don't know the exact value. However, let me note that 3 + 2√2 is equal to (1 + √2)^2. Let me check: (1 + √2)^2 = 1 + 2√2 + 2 = 3 + 2√2. Yes! Therefore, tan α = (1 + √2)^2. Let me think about tan(π/8) and tan(3π/8). Wait, tan(π/8) = √2 - 1 ≈ 0.414, tan(3π/8) = √2 + 1 ≈ 2.414. Let me compute tan(π/4 + π/8) = tan(3π/8) = √2 + 1. Similarly, tan(π/4 + 3π/8) = tan(5π/8) = -tan(3π/8) = -√2 - 1. Not sure. Alternatively, let me consider that tan(π/2 - α) = cot α = 3 + 2√2. Therefore, α = arccot(3 + 2√2). But cot α = 3 + 2√2, so tan α = 1/(3 + 2√2). Let me rationalize 1/(3 + 2√2) = (3 - 2√2)/[(3 + 2√2)(3 - 2√2)] = (3 - 2√2)/(9 - 8) = 3 - 2√2. Therefore, tan α = 3 - 2√2. Therefore, α = arctan(3 - 2√2). Let me compute 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172. Therefore, α ≈ arctan(0.172) ≈ 0.17 radians. Therefore, π/2 - α ≈ π/2 - 0.17 ≈ 1.47 radians. Therefore, tan(π/2 - α) = cot α = 3 + 2√2. Therefore, π/2 - α = arctan(3 + 2√2). Therefore, arctan(3 + 2√2) = π/2 - arctan(3 - 2√2). Therefore, going back, tan⁻¹(-x) = tan⁻¹(3 + 2√2) = π/2 - arctan(3 - 2√2). Therefore, 2 tan⁻¹(-x) = π - 2 arctan(3 - 2√2). Therefore, π - 2 tan⁻¹(-x) = 2 arctan(3 - 2√2). Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2 arctan(3 - 2√2)). Let me compute sin(2 arctan(3 - 2√2)). Let me denote β = arctan(3 - 2√2). Then tan β = 3 - 2√2. Then sin(2β) = 2 tan β / (1 + tan² β). Let me compute tan β = 3 - 2√2. Then tan² β = (3 - 2√2)^2 = 9 - 12√2 + 8 = 17 - 12√2. Therefore, 1 + tan² β = 18 - 12√2. Therefore, sin(2β) = 2*(3 - 2√2)/(18 - 12√2). Let me compute this:

Multiply numerator and denominator by the conjugate of the denominator's denominator? Wait, let me rationalize:

Numerator: 2*(3 - 2√2) = 6 - 4√2.

Denominator: 18 - 12√2 = 6*(3 - 2√2).

Therefore, sin(2β) = (6 - 4√2)/[6*(3 - 2√2)] = [2*(3 - 2√2)]/[6*(3 - 2√2)] = 2/6 = 1/3. Therefore, sin(2β) = 1/3. Therefore, sin(π - 2 tan⁻¹(-x)) = 1/3. Therefore, this is correct. Therefore, even though numerically I thought that sin(1.55) ≈ 0.999, which conflicts, but actually, the angle π - 2 tan⁻¹(-x) is equal to 2β, which is 2 arctan(3 - 2√2). Let me compute 2β. Since β = arctan(3 - 2√2) ≈ arctan(0.172) ≈ 0.17 radians. Therefore, 2β ≈ 0.34 radians. Then sin(0.34) ≈ 0.333, which is 1/3. Therefore, my previous numerical calculation was wrong because I miscalculated the angle. Earlier, I thought that π - 2 tan⁻¹(-x) ≈ 1.55 radians, but actually, π - 2 tan⁻¹(-x) = 2β ≈ 0.34 radians. Wait, how?

Wait, earlier I said that tan⁻¹(-x) = π/2 - β, where β = arctan(3 - 2√2). Therefore, 2 tan⁻¹(-x) = π - 2β. Therefore, π - 2 tan⁻¹(-x) = 2β. Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2β) = 1/3. Therefore, the angle π - 2 tan⁻¹(-x) is 2β ≈ 0.34 radians, not 1.55 radians. Therefore, my previous mistake was in calculating π - 2 tan⁻¹(-x). Let me recalculate:

For x = -3 - 2√2 ≈ -5.828, tan⁻¹(-x) = tan⁻¹(3 + 2√2) ≈ π/2 - β ≈ π/2 - 0.17 ≈ 1.47 radians. Therefore, 2 tan⁻¹(-x) ≈ 2.94 radians. Therefore, π - 2 tan⁻¹(-x) ≈ π - 2.94 ≈ 0.2 radians. Wait, π ≈ 3.14, so 3.14 - 2.94 ≈ 0.2 radians. Therefore, sin(0.2) ≈ 0.198, which is not 1/3. Wait, this is conflicting with the previous result. What is happening here?

Wait, there must be confusion in the angle calculations. Let me try again. Let me denote β = arctan(3 - 2√2). Then tan β = 3 - 2√2. Then, as we saw, sin(2β) = 1/3. Therefore, 2β = arcsin(1/3) or π - arcsin(1/3). Let me compute arcsin(1/3) ≈ 0.3398 radians. Therefore, 2β ≈ 0.3398 radians, so β ≈ 0.1699 radians. Therefore, tan β ≈ tan(0.1699) ≈ 0.172, which matches 3 - 2√2 ≈ 0.172. Therefore, β ≈ 0.17 radians. Therefore, tan⁻¹(-x) = π/2 - β ≈ π/2 - 0.17 ≈ 1.57 - 0.17 ≈ 1.4 radians. Therefore, 2 tan⁻¹(-x) ≈ 2.8 radians. Therefore, π - 2 tan⁻¹(-x) ≈ π - 2.8 ≈ 0.34 radians. Therefore, sin(π - 2 tan⁻¹(-x)) = sin(0.34) ≈ 0.333, which is 1/3. Therefore, the angle π - 2 tan⁻¹(-x) is approximately 0.34 radians, not 1.55 radians. My previous mistake was in calculating π - 2 tan⁻¹(-x). I incorrectly thought that tan⁻¹(-x) was -1.42 radians, but actually tan⁻¹(-x) = tan⁻¹(3 + 2√2) ≈ π/2 - β ≈ 1.4 radians, not -1.42 radians. Wait, no. Wait, x = -3 - 2√2, so -x = 3 + 2√2. Therefore, tan⁻¹(-x) = tan⁻¹(3 + 2√2). Since 3 + 2√2 is positive, tan⁻¹(3 + 2√2) is in (0, π/2). Wait, no! Wait, tan⁻¹(y) for y > 0 is in (0, π/2). Therefore, tan⁻¹(3 + 2√2) is in (0, π/2). Therefore, tan⁻¹(-x) is in (0, π/2). Therefore, π - 2 tan⁻¹(-x) is π - 2*(something in (0, π/2)). Therefore, π - 2*(something in (0, π/2)) is in (π - π, π - 0) = (0, π). Therefore, sin(π - 2 tan⁻¹(-x)) is in (0, 1]. Therefore, sin(π - 2 tan⁻¹(-x)) = 1/3 is valid. Therefore, my previous calculation where I thought tan⁻¹(x) ≈ -1.42 radians was wrong. Wait, tan⁻¹(x) where x = -3 - 2√2 is tan⁻¹(-5.828), which is in (-π/2, 0). Therefore, tan⁻¹(x) ≈ -1.42 radians. But cot⁻¹(1/x) = cot⁻¹(-1/5.828) = π - cot⁻¹(1/5.828) ≈ π - tan⁻¹(5.828) ≈ π - 1.42 ≈ 1.72 radians. Therefore, adding tan⁻¹(x) + cot⁻¹(1/x) ≈ -1.42 + 1.72 ≈ 0.3 radians. Therefore, sin(0.3) ≈ 0.295 ≈ 1/3. Therefore, this is correct. Therefore, my previous mistake was in calculating π - 2 tan⁻¹(-x) as 0.34 radians, which is the same as tan⁻¹(x) + cot⁻¹(1/x) ≈ 0.3 radians. Therefore, everything checks out. Therefore, all four solutions are valid.

Therefore, the solutions are x = 3 + 2√2, x = 3 - 2√2, x = -3 + 2√2, and x = -3 - 2√2.

But let me check if there are any other solutions. For example, when solving the equation sin(θ) = 1/3, θ can be in multiple quadrants. However, in our case, θ = tan⁻¹(x) + cot⁻¹(1/x). For x > 0, θ = 2 tan⁻¹(x), which is between 0 and π. For x < 0, θ = π - 2 tan⁻¹(-x), which is between 0 and π. Therefore, sin(θ) = 1/3 has two solutions in (0, π): arcsin(1/3) and π - arcsin(1/3). Therefore, for each case (x > 0 and x < 0), we have two solutions. Therefore, total four solutions. Therefore, our four solutions are correct.

Therefore, the final answer is all four solutions: 3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

But let me check if these can be simplified or written differently. For example, 3 - 2√2 is already simplified, and -3 + 2√2 is also simplified. Similarly for the others. Therefore, the solutions are:

x = 3 + 2√2,

x = 3 - 2√2,

x = -3 + 2√2,

x = -3 - 2√2.

Therefore, the answer should be these four values.

But let me check if there's any mistake in the quadratic equations. For x > 0, we had 2x/(1 + x²) = 1/3 leading to x² - 6x + 1 = 0, solutions 3 ± 2√2. For x < 0, we had -2x/(1 + x²) = 1/3 leading to x² + 6x + 1 = 0, solutions -3 ± 2√2. Therefore, yes, that's correct. Therefore, the four solutions are correct.

Therefore, the solutions are:

3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

But let me check if these are all distinct. Yes, as we saw numerically, they are four distinct real numbers.

Therefore, the final answer is these four values.

**Final Answer**
The solutions are $\boxed{-3 - 2\sqrt{2}}$, $\boxed{-3 + 2\sqrt{2}}$, $\boxed{3 - 2\sqrt{2}}$, and $\boxed{3 + 2\sqrt{2}}$.

To solve the equation:

$$
\sin \left( \tan^{-1} (x) + \cot^{-1} \left( \frac{1}{x} \right) \right) = \frac{1}{3},
$$

we begin by analyzing the expression inside the sine function.

---

### Step 1: Simplify the Argument of the Sine Function

We use the identity:

$$
\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right),
$$

which holds for $ y > 0 $. Therefore, for $ x > 0 $, we have:

$$
\cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(x).
$$

So the expression becomes:

$$
\tan^{-1}(x) + \cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(x) + \tan^{-1}(x) = 2 \tan^{-1}(x).
$$

Thus, the equation becomes:

$$
\sin\left(2 \tan^{-1}(x)\right) = \frac{1}{3}.
$$

---

### Step 2: Use the Double Angle Identity

Recall the identity:

$$
\sin(2\theta) = 2 \sin(\theta) \cos(\theta).
$$

Let $ \theta = \tan^{-1}(x) $. Then:

- $ \sin(\theta) = \frac{x}{\sqrt{1 + x^2}} $,
- $ \cos(\theta) = \frac{1}{\sqrt{1 + x^2}} $.

So:

$$
\sin(2\theta) = 2 \cdot \frac{x}{\sqrt{1 + x^2}} \cdot \frac{1}{\sqrt{1 + x^2}} = \frac{2x}{1 + x^2}.
$$

Therefore, the equation becomes:

$$
\frac{2x}{1 + x^2} = \frac{1}{3}.
$$

---

### Step 3: Solve the Equation

Cross-multiplying:

$$
6x = 1 + x^2 \quad \Rightarrow \quad x^2 - 6x + 1 = 0.
$$

Solving this quadratic:

$$
x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}.
$$

So, the two positive solutions are:

$$
x = 3 + 2\sqrt{2}, \quad x = 3 - 2\sqrt{2}.
$$

---

### Step 4: Consider Negative Values of $ x $

Now consider $ x < 0 $. In this case, the identity $ \cot^{-1}(1/x) = \tan^{-1}(x) $ does **not** hold, because $ \cot^{-1}(1/x) $ for negative $ x $ lies in the second quadrant, while $ \tan^{-1}(x) $ lies in the fourth.

Instead, we use the identity:

$$
\tan^{-1}(x) + \cot^{-1}\left(\frac{1}{x}\right) = \pi - 2 \tan^{-1}(-x),
$$

for $ x < 0 $. Then:

$$
\sin\left(\pi - 2 \tan^{-1}(-x)\right) = \sin\left(2 \tan^{-1}(-x)\right) = \frac{1}{3}.
$$

Let $ y = -x $, so $ y > 0 $. Then:

$$
\sin\left(2 \tan^{-1}(y)\right) = \frac{1}{3} \quad \Rightarrow \quad \frac{2y}{1 + y^2} = \frac{1}{3}.
$$

This is the same equation as before, leading to:

$$
y^2 - 6y + 1 = 0 \quad \Rightarrow \quad y = 3 \pm 2\sqrt{2}.
$$

Since $ y = -x $, we get:

$$
x = -3 \pm 2\sqrt{2}.
$$

---

### Step 5: Final Answer

Combining all four solutions, we have:

$$
x = 3 + 2\sqrt{2}, \quad x = 3 - 2\sqrt{2}, \quad x = -3 + 2\sqrt{2}, \quad x = -3 - 2\sqrt{2}.
$$

All four values are valid and distinct.

---

### Final Answer

$$
\boxed{-3 - 2\sqrt{2}}, \quad \boxed{-3 + 2\sqrt{2}}, \quad \boxed{3 - 2\sqrt{2}}, \quad \boxed{3 + 2\sqrt{2}}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation:

$$
\sin \left( \tan^{-1} (x) + \cot^{-1} \left( \frac{1}{x} \right) \right) = \frac{1}{3}.
$$

Hmm, let me start by recalling some trigonometric identities and inverse trigonometric functions properties. Maybe I can simplify the expression inside the sine function first. Let me denote:

Let’s let θ = tan⁻¹(x). Then tan θ = x. Also, cot⁻¹(1/x) is another term. I remember that cot⁻¹(y) is equal to tan⁻¹(1/y), right? Wait, is that true?

Wait, let me check. The cotangent is the reciprocal of tangent, so cot(α) = 1/tan(α). Therefore, if cot⁻¹(y) is the angle whose cotangent is y, then that angle is equal to tan⁻¹(1/y). So yes, cot⁻¹(y) = tan⁻¹(1/y). Therefore, cot⁻¹(1/x) = tan⁻¹(x). Wait, that would mean that cot⁻¹(1/x) is tan⁻¹(x). But if θ = tan⁻¹(x), then cot⁻¹(1/x) is also θ? Let me verify with an example. Suppose x = 1. Then tan⁻¹(1) is π/4, and cot⁻¹(1/1) = cot⁻¹(1) which is also π/4. So yes, that seems to hold. So in general, cot⁻¹(1/x) = tan⁻¹(x). Therefore, the argument inside the sine function is tan⁻¹(x) + tan⁻¹(x) = 2 tan⁻¹(x). Therefore, the equation simplifies to:

sin(2 tan⁻¹(x)) = 1/3.

Wait, is that correct? Let me check again. If cot⁻¹(1/x) = tan⁻¹(x), then adding tan⁻¹(x) + cot⁻¹(1/x) is tan⁻¹(x) + tan⁻¹(x) = 2 tan⁻¹(x). Therefore, the equation becomes sin(2 tan⁻¹(x)) = 1/3. That seems right. So maybe I can use the double angle formula for sine here. Let me recall that sin(2α) = 2 sin α cos α. So if I let α = tan⁻¹(x), then sin(2α) = 2 sin α cos α. But since α = tan⁻¹(x), then we can think of a right triangle where the opposite side is x and adjacent side is 1, so hypotenuse is √(1 + x²). Therefore, sin α = x / √(1 + x²) and cos α = 1 / √(1 + x²). Therefore, sin(2α) = 2*(x / √(1 + x²))*(1 / √(1 + x²)) = 2x / (1 + x²). Therefore, the equation becomes:

2x / (1 + x²) = 1/3.

So now, solving 2x / (1 + x²) = 1/3. Let me cross multiply:

2x * 3 = 1 * (1 + x²)

Which is 6x = 1 + x²

Rearranging terms:

x² - 6x + 1 = 0

Wait, hold on. Let me check that again. If 6x = 1 + x², then bringing all terms to one side:

x² - 6x + 1 = 0? Wait, no. If 6x = 1 + x², then subtract 6x from both sides:

0 = x² - 6x + 1. Yes, that's correct. So quadratic equation x² - 6x + 1 = 0. Let me solve this quadratic equation. Using quadratic formula:

x = [6 ± √(36 - 4*1*1)] / 2 = [6 ± √(32)] / 2 = [6 ± 4√2]/2 = 3 ± 2√2.

So the solutions are x = 3 + 2√2 and x = 3 - 2√2. Wait, but I need to check if these solutions are valid in the original equation. Because sometimes when dealing with inverse trigonometric functions, there might be restrictions on the domain or some extraneous solutions introduced.

But let me think. The original equation involves tan⁻¹(x) and cot⁻¹(1/x). The domain of tan⁻¹(x) is all real numbers, and cot⁻¹(1/x) requires that 1/x is defined, so x ≠ 0. Therefore, the domain of the original equation is x ≠ 0. So our solutions 3 + 2√2 and 3 - 2√2 are both non-zero, so they are in the domain. But maybe there's something else?

Wait, but let me check if there are any restrictions on the range of the inverse functions. For example, tan⁻¹(x) gives values between -π/2 and π/2, and cot⁻¹(1/x) gives values between 0 and π. However, when we added them together, we need to make sure that the sum is within the domain where sine is defined, which it always is. But maybe there's a problem with the identity I used earlier?

Wait, earlier I said that cot⁻¹(1/x) = tan⁻¹(x). Let me check if that's always true. Let me consider x positive first. If x is positive, then tan⁻¹(x) is between 0 and π/2, and cot⁻¹(1/x) is also between 0 and π/2. Since cot⁻¹(y) = tan⁻¹(1/y) for y > 0, so if y = 1/x, then cot⁻¹(1/x) = tan⁻¹(x). So that's correct for x > 0. What if x is negative?

Suppose x is negative. Let me take x = -a where a > 0. Then tan⁻¹(x) = tan⁻¹(-a) = -tan⁻¹(a). On the other hand, cot⁻¹(1/x) = cot⁻¹(-1/a). But cot⁻¹(y) for negative y. Wait, the range of cot⁻¹(y) is typically (0, π). So cot⁻¹(-1/a) would be π - cot⁻¹(1/a). Because cotangent is negative in the second quadrant. Let me recall that cot⁻¹(y) is defined as the angle in (0, π) whose cotangent is y. So if y is negative, then cot⁻¹(y) is in (π/2, π). So for example, cot⁻¹(-1/a) = π - cot⁻¹(1/a). But cot⁻¹(1/a) is tan⁻¹(a). Therefore, cot⁻¹(-1/a) = π - tan⁻¹(a). Therefore, cot⁻¹(1/x) when x is negative is π - tan⁻¹(a) = π - tan⁻¹(-x). Therefore, in that case, tan⁻¹(x) + cot⁻¹(1/x) = tan⁻¹(-a) + π - tan⁻¹(a) = -tan⁻¹(a) + π - tan⁻¹(a) = π - 2 tan⁻¹(a). But if we consider the original expression, when x is negative, tan⁻¹(x) + cot⁻¹(1/x) is π - 2 tan⁻¹(-x). Wait, but in our previous step, we assumed that cot⁻¹(1/x) = tan⁻¹(x), which is only true when x is positive. For negative x, that identity doesn't hold. Therefore, my earlier conclusion that tan⁻¹(x) + cot⁻¹(1/x) = 2 tan⁻¹(x) is only valid when x is positive. If x is negative, then the sum is π - 2 tan⁻¹(-x). Therefore, we need to check if x is positive or negative.

Therefore, perhaps my initial simplification was only valid for x > 0. Therefore, maybe I need to consider two cases: x > 0 and x < 0.

So let me re-examine the original problem. Let me first check if x is positive or negative.

Case 1: x > 0.

In this case, as we saw, cot⁻¹(1/x) = tan⁻¹(x). Therefore, tan⁻¹(x) + cot⁻¹(1/x) = 2 tan⁻¹(x). Therefore, sin(2 tan⁻¹(x)) = 1/3. Which led us to the equation 2x/(1 + x²) = 1/3, which gave solutions x = 3 ± 2√2. However, since x is positive, we need to check which of these solutions are positive. 3 + 2√2 is definitely positive. 3 - 2√2 is approximately 3 - 2.828 ≈ 0.172, which is also positive. Therefore, both solutions are positive. Therefore, both are valid in this case.

Case 2: x < 0.

In this case, as we saw, tan⁻¹(x) + cot⁻¹(1/x) = π - 2 tan⁻¹(-x). Let me denote y = -x, so y > 0. Then tan⁻¹(x) + cot⁻¹(1/x) = π - 2 tan⁻¹(y). Therefore, the original equation becomes:

sin(π - 2 tan⁻¹(y)) = 1/3.

But sin(π - θ) = sin θ. Therefore, sin(π - 2 tan⁻¹(y)) = sin(2 tan⁻¹(y)) = 1/3. Therefore, this reduces to the same equation as in Case 1: sin(2 tan⁻¹(y)) = 1/3. Which again leads to 2y/(1 + y²) = 1/3. Therefore, solving for y gives y = 3 ± 2√2. But since y = -x and x < 0, then y > 0. Therefore, y can be 3 + 2√2 or 3 - 2√2. Therefore, x = -y, so x = -3 - 2√2 or x = -3 + 2√2. However, we need to check if these solutions are valid. Let me check if they satisfy the original equation.

But wait, let me think again. Since we have sin(π - 2 tan⁻¹(y)) = sin(2 tan⁻¹(y)) = 1/3. Therefore, the equation reduces to the same equation as in Case 1. Therefore, solving for y gives the same solutions. Therefore, in Case 2, we have x = -y, so x = - (3 + 2√2) and x = - (3 - 2√2). However, we need to check if these solutions are valid. Let me check if they satisfy the original equation.

But let me check if there's any problem with the steps. For example, when we have sin(π - θ) = sin θ, which is true. Therefore, if we have sin(π - 2 tan⁻¹(y)) = 1/3, then sin(2 tan⁻¹(y)) = 1/3. Therefore, solving for y gives the same equation as before. Therefore, the solutions for y are 3 ± 2√2, so x = - (3 ± 2√2). Therefore, in total, we have four solutions? Wait, but in Case 1, x > 0 gives two solutions, and in Case 2, x < 0 gives two solutions. But let me check if all four solutions are valid.

Wait, but let me check if when x is negative, the original equation is valid. Let me take x = - (3 - 2√2). Let me compute 3 - 2√2. Since √2 ≈ 1.414, so 2√2 ≈ 2.828. Therefore, 3 - 2.828 ≈ 0.172. Therefore, x ≈ -0.172. Let me check if this satisfies the original equation.

Alternatively, maybe there's a mistake here. Let me think again. When we have x negative, we have tan⁻¹(x) + cot⁻¹(1/x) = π - 2 tan⁻¹(-x). Then sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x))? Wait, no. Wait, sin(π - θ) = sin θ. Therefore, sin(π - 2 tan⁻¹(y)) = sin(2 tan⁻¹(y)). Therefore, the equation becomes sin(2 tan⁻¹(y)) = 1/3. Therefore, solving for y gives the same equation as before. Therefore, the solutions for y are 3 ± 2√2, so x = -y. Therefore, x = -3 - 2√2 and x = -3 + 2√2. However, we need to check if these solutions are valid. Let me check if they satisfy the original equation.

But let me check if there's a problem with the range of the inverse functions. For example, when x is negative, tan⁻¹(x) is between -π/2 and 0, and cot⁻¹(1/x) is between π/2 and π. Therefore, their sum is between 0 and π. Therefore, sin of that sum is between 0 and 1. Since 1/3 is positive, that's okay. Therefore, the solutions should be valid. Therefore, in total, there are four solutions? Wait, but when I solved the equation, I had x² - 6x + 1 = 0, which gave two solutions. But if we consider both positive and negative x, then we have four solutions? Wait, no. Wait, the equation 2x/(1 + x²) = 1/3 led to x² - 6x + 1 = 0, which has two solutions. However, when we considered x negative, we found that the equation reduces to the same equation for y, which is positive. Therefore, the solutions for x negative are - (3 + 2√2) and - (3 - 2√2). Therefore, in total, there are four solutions? Wait, but the equation 2x/(1 + x²) = 1/3 is a quadratic equation, which can have at most two real solutions. Therefore, how come we have four solutions? There must be a mistake here.

Wait, no. Wait, when I considered x negative, I transformed the original equation into the same equation as for x positive. Therefore, solving 2x/(1 + x²) = 1/3 gives two solutions, but when x is negative, we have to check if they satisfy the original equation. Wait, but if we have x negative, then the equation sin(π - 2 tan⁻¹(y)) = 1/3 becomes sin(2 tan⁻¹(y)) = 1/3, which is the same equation as before. Therefore, solving for y gives the same solutions. Therefore, the solutions for x negative are - (3 + 2√2) and - (3 - 2√2). However, when we plug these into the equation 2x/(1 + x²) = 1/3, do they satisfy?

Let me check. Let me take x = - (3 + 2√2). Let me compute 2x/(1 + x²). Let me compute x²:

x = - (3 + 2√2). Then x² = (3 + 2√2)^2 = 9 + 12√2 + 8 = 17 + 12√2.

Then 1 + x² = 18 + 12√2.

Then 2x = -2*(3 + 2√2) = -6 - 4√2.

Therefore, 2x/(1 + x²) = (-6 - 4√2)/(18 + 12√2). Let me rationalize the denominator or simplify:

Factor numerator and denominator:

Numerator: -2*(3 + 2√2)

Denominator: 6*(3 + 2√2)

Therefore, the ratio is (-2/6) = -1/3. But the original equation is sin(...) = 1/3. However, we have 2x/(1 + x²) = -1/3. But earlier, we had that sin(2 tan⁻¹(x)) = 2x/(1 + x²). However, when x is negative, we had that the original equation reduces to sin(π - 2 tan⁻¹(y)) = 1/3, which is equal to sin(2 tan⁻¹(y)) = 1/3. Therefore, in this case, the value of 2x/(1 + x²) is -1/3, but the sine of the angle is 1/3. Therefore, there is a contradiction here. Therefore, my previous reasoning was flawed.

Wait, so where is the mistake? Let me retrace. When x is negative, we have:

Original expression inside sine: tan⁻¹(x) + cot⁻¹(1/x) = π - 2 tan⁻¹(-x). Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x))? Wait, no. Wait, sin(π - θ) = sin θ. Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x)). Wait, but tan⁻¹(-x) is equal to - tan⁻¹(x). Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x)) = sin(-2 tan⁻¹(x)) = - sin(2 tan⁻¹(x)). Therefore, the original equation becomes - sin(2 tan⁻¹(x)) = 1/3. Therefore, sin(2 tan⁻¹(x)) = -1/3. Therefore, in this case, the equation is not 2x/(1 + x²) = 1/3, but rather 2x/(1 + x²) = -1/3. Therefore, my mistake was in the previous step where I thought that sin(π - 2 tan⁻¹(y)) = sin(2 tan⁻¹(y)), but actually, it's sin(π - θ) = sin θ, but θ here is 2 tan⁻¹(y). Wait, no. Wait, if θ = 2 tan⁻¹(y), then sin(π - θ) = sin θ. Therefore, sin(π - θ) = sin θ. Therefore, if the original equation is sin(π - θ) = 1/3, then sin θ = 1/3. Therefore, θ = arcsin(1/3) or π - arcsin(1/3). Therefore, in this case, θ = 2 tan⁻¹(y). Therefore, 2 tan⁻¹(y) = arcsin(1/3) or π - arcsin(1/3). Therefore, solving for y would give different solutions. Therefore, my previous conclusion that it reduces to the same equation was wrong. Therefore, I need to correct this.

Therefore, going back. For x < 0, we have:

Original equation: sin(π - 2 tan⁻¹(-x)) = 1/3. Let me denote θ = 2 tan⁻¹(-x). Then sin(π - θ) = sin θ = 1/3. Therefore, sin θ = 1/3. Therefore, θ = arcsin(1/3) or π - arcsin(1/3). Therefore, 2 tan⁻¹(-x) = arcsin(1/3) or π - arcsin(1/3). Therefore, tan⁻¹(-x) = (1/2) arcsin(1/3) or (π - arcsin(1/3))/2. Therefore, -x = tan[(1/2) arcsin(1/3)] or tan[(π - arcsin(1/3))/2]. Therefore, solving for x would give different expressions. However, this seems complicated. Alternatively, maybe I can relate this back to the equation 2x/(1 + x²) = -1/3. Because earlier, for x positive, we had sin(2 tan⁻¹(x)) = 2x/(1 + x²). For x negative, we have sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x)) = - sin(2 tan⁻¹(x)) = - [2x/(1 + x²)]? Wait, let me check.

If x is negative, then tan⁻¹(x) = - tan⁻¹(-x). Let me denote α = tan⁻¹(-x), which is positive because -x is positive. Then tan α = -x. Therefore, sin(2 tan⁻¹(x)) = sin(-2α) = - sin(2α). But 2α = 2 tan⁻¹(-x). Therefore, sin(π - 2α) = sin(2α) = 1/3. Therefore, sin(2α) = 1/3. Therefore, sin(2α) = 1/3. But 2α = 2 tan⁻¹(-x). Therefore, sin(2 tan⁻¹(-x)) = 1/3. But sin(2 tan⁻¹(-x)) = 2*(-x)/(1 + (-x)^2) = -2x/(1 + x²). Therefore, -2x/(1 + x²) = 1/3. Therefore, solving -2x/(1 + x²) = 1/3. Which leads to -2x = (1 + x²)/3 => -6x = 1 + x² => x² + 6x + 1 = 0. Therefore, solving x² + 6x + 1 = 0. Therefore, x = [-6 ± √(36 - 4)] / 2 = [-6 ± √32]/2 = [-6 ± 4√2]/2 = -3 ± 2√2. Therefore, x = -3 + 2√2 or x = -3 - 2√2. Therefore, these are the solutions for x < 0. Therefore, in total, the solutions are:

From x > 0: 3 + 2√2 and 3 - 2√2.

From x < 0: -3 + 2√2 and -3 - 2√2.

But wait, let me check if these solutions are valid. Let me check for x = -3 + 2√2. Let me compute approximate values. √2 ≈ 1.414, so 2√2 ≈ 2.828. Therefore, -3 + 2.828 ≈ -0.172. Similarly, -3 - 2√2 ≈ -5.828. Let me check if these satisfy the original equation.

First, take x = -3 + 2√2 ≈ -0.172. Let me compute tan⁻¹(x) + cot⁻¹(1/x). Let me compute tan⁻¹(x). Since x is negative, tan⁻¹(x) is negative. Let me compute cot⁻¹(1/x). Since x is negative, 1/x is negative. cot⁻¹(1/x) is in (π/2, π). Let me compute tan⁻¹(x) + cot⁻¹(1/x). Let me denote θ = tan⁻¹(x). Then cot⁻¹(1/x) = π - tan⁻¹(-1/x) = π - tan⁻¹(1/|x|). Wait, but maybe it's easier to compute numerically.

Alternatively, since we derived that for x negative, the equation reduces to -2x/(1 + x²) = 1/3. Let me check for x = -3 + 2√2. Let me compute -2x/(1 + x²):

First, compute x = -3 + 2√2. Let me compute x²:

x = -3 + 2√2. Then x² = (-3)^2 + (2√2)^2 - 2*3*2√2 = 9 + 8 - 12√2 = 17 - 12√2.

Then 1 + x² = 1 + 17 - 12√2 = 18 - 12√2.

Then -2x = -2*(-3 + 2√2) = 6 - 4√2.

Therefore, -2x/(1 + x²) = (6 - 4√2)/(18 - 12√2). Let me rationalize the denominator:

Multiply numerator and denominator by (18 + 12√2):

Numerator: (6 - 4√2)(18 + 12√2)

Denominator: (18 - 12√2)(18 + 12√2) = 18² - (12√2)² = 324 - 288 = 36.

Compute numerator:

First, expand (6)(18) + 6*(12√2) - 4√2*(18) - 4√2*(12√2)

= 108 + 72√2 - 72√2 - 48*(√2)^2

Simplify:

108 + (72√2 - 72√2) - 48*2

= 108 - 96 = 12.

Therefore, numerator is 12, denominator is 36. Therefore, 12/36 = 1/3. Therefore, -2x/(1 + x²) = 1/3. Therefore, this satisfies the equation. Therefore, x = -3 + 2√2 is a valid solution. Similarly, for x = -3 - 2√2, let me check:

x = -3 - 2√2. Compute -2x/(1 + x²):

First, x = -3 - 2√2. Then x² = (-3 - 2√2)^2 = 9 + 12√2 + 8 = 17 + 12√2.

Then 1 + x² = 18 + 12√2.

-2x = -2*(-3 - 2√2) = 6 + 4√2.

Therefore, -2x/(1 + x²) = (6 + 4√2)/(18 + 12√2). Again, rationalize denominator by multiplying numerator and denominator by (18 - 12√2):

Numerator: (6 + 4√2)(18 - 12√2)

Denominator: same as before, 36.

Compute numerator:

6*18 + 6*(-12√2) + 4√2*18 + 4√2*(-12√2)

= 108 - 72√2 + 72√2 - 48*(√2)^2

= 108 - 0 - 48*2 = 108 - 96 = 12.

Therefore, 12/36 = 1/3. Therefore, this also satisfies the equation. Therefore, both x = -3 + 2√2 and x = -3 - 2√2 are valid solutions.

Therefore, in total, the four solutions are:

x = 3 + 2√2,

x = 3 - 2√2,

x = -3 + 2√2,

x = -3 - 2√2.

But wait, let me check if these are all distinct. Let me compute approximate values:

3 + 2√2 ≈ 3 + 2.828 ≈ 5.828,

3 - 2√2 ≈ 3 - 2.828 ≈ 0.172,

-3 + 2√2 ≈ -3 + 2.828 ≈ -0.172,

-3 - 2√2 ≈ -3 - 2.828 ≈ -5.828.

So all four solutions are distinct. Therefore, the original equation has four solutions. However, let me check if there are any restrictions. For example, when x = 0, the original equation is undefined, but our solutions don't include 0. Also, for all other x, the functions are defined. Therefore, all four solutions are valid.

But wait, let me check with the original equation. Let me take x = 3 - 2√2 ≈ 0.172. Let me compute tan⁻¹(x) + cot⁻¹(1/x). Let me compute tan⁻¹(x) ≈ tan⁻¹(0.172) ≈ 0.17 radians (approx). Then 1/x ≈ 5.828, so cot⁻¹(5.828) ≈ tan⁻¹(1/5.828) ≈ tan⁻¹(0.172) ≈ 0.17 radians. Therefore, adding them together gives approx 0.34 radians. Then sin(0.34) ≈ 0.333, which is approx 1/3. Therefore, that works. Similarly, for x = -0.172, tan⁻¹(x) ≈ -0.17 radians, and cot⁻¹(1/x) = cot⁻¹(-5.828) which is π - tan⁻¹(5.828) ≈ π - 1.42 radians ≈ 1.72 radians. Adding tan⁻¹(x) + cot⁻¹(1/x) ≈ -0.17 + 1.72 ≈ 1.55 radians. Then sin(1.55) ≈ sin(π - 1.55) ≈ sin(1.59) ≈ 0.999? Wait, no. Wait, sin(1.55) is approx sin(π - 1.55) = sin(1.59) ≈ 0.999? Wait, no. Wait, π is approx 3.14, so π - 1.55 ≈ 1.59 radians. Wait, sin(1.55) is approx sin(π - 1.55) = sin(1.59). Wait, but sin(1.55) ≈ sin(88.8 degrees) ≈ 0.999. Wait, but 1.55 radians is approx 88.8 degrees? Wait, π/2 is approx 1.57 radians, so 1.55 radians is slightly less than π/2, so sin(1.55) ≈ sin(π/2 - 0.02) ≈ cos(0.02) ≈ 0.9998. But according to our equation, it should be 1/3 ≈ 0.333. Wait, this seems contradictory. What's wrong here?

Wait, there must be a miscalculation here. Let me check again. If x = -0.172, then tan⁻¹(x) ≈ -0.17 radians. Then cot⁻¹(1/x) = cot⁻¹(-5.828). Since cot⁻¹(y) for negative y is π - cot⁻¹(|y|). So cot⁻¹(-5.828) = π - cot⁻¹(5.828). But cot⁻¹(5.828) is tan⁻¹(1/5.828) ≈ tan⁻¹(0.172) ≈ 0.17 radians. Therefore, cot⁻¹(-5.828) ≈ π - 0.17 ≈ 3.14 - 0.17 ≈ 2.97 radians. Therefore, tan⁻¹(x) + cot⁻¹(1/x) ≈ -0.17 + 2.97 ≈ 2.8 radians. Then sin(2.8 radians). Let me compute sin(2.8). 2.8 radians is approx 160 degrees (since π radians is 180 degrees, so 2.8 radians ≈ 160 degrees). sin(160 degrees) = sin(20 degrees) ≈ 0.342, which is approx 1/3 (≈ 0.333). Therefore, that works. Therefore, sin(2.8) ≈ 0.342 ≈ 1/3. Therefore, that's correct. Therefore, the solution x = -0.172 is valid. Similarly, for x = -5.828, let me check. tan⁻¹(x) ≈ tan⁻¹(-5.828) ≈ -1.42 radians. cot⁻¹(1/x) = cot⁻¹(-1/5.828) = cot⁻¹(-0.172) = π - cot⁻¹(0.172) ≈ π - 0.17 ≈ 2.97 radians. Therefore, adding tan⁻¹(x) + cot⁻¹(1/x) ≈ -1.42 + 2.97 ≈ 1.55 radians. Then sin(1.55) ≈ sin(π - 1.55) = sin(1.59) ≈ 0.999? Wait, no. Wait, sin(1.55) is approx sin(88.8 degrees) ≈ 0.999, but according to our equation, it should be 1/3. Wait, this is conflicting. Wait, but earlier we had that for x negative, the equation reduces to -2x/(1 + x²) = 1/3. Let me check for x = -3 - 2√2 ≈ -5.828. Let me compute -2x/(1 + x²). As we did before, it equals 1/3. Therefore, according to our previous derivation, sin(π - 2 tan⁻¹(-x)) = 1/3. Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2 tan⁻¹(-x)) = 1/3. Wait, but earlier I thought that sin(π - θ) = sin θ, so sin(π - θ) = sin θ. Therefore, if θ = 2 tan⁻¹(-x), then sin(π - θ) = sin θ = 1/3. Therefore, sin(θ) = 1/3. Therefore, θ = arcsin(1/3) or π - arcsin(1/3). Therefore, 2 tan⁻¹(-x) = arcsin(1/3) or π - arcsin(1/3). Therefore, tan⁻¹(-x) = (1/2) arcsin(1/3) or (π - arcsin(1/3))/2. Therefore, -x = tan[(1/2) arcsin(1/3)] or tan[(π - arcsin(1/3))/2]. Therefore, solving for x gives us the solutions we found. Therefore, even though when I compute numerically, I get sin(1.55) ≈ 0.999, which conflicts with 1/3, there must be an error in my numerical estimation. Wait, let me compute 1.55 radians. Let me convert 1.55 radians to degrees: 1.55 * (180/π) ≈ 1.55 * 57.3 ≈ 88.8 degrees. Wait, sin(88.8 degrees) is indeed approximately 0.999. But according to our equation, it should be 1/3. Therefore, there is a contradiction here. What is wrong?

Wait, let me check again. For x = -3 - 2√2 ≈ -5.828. Let me compute tan⁻¹(x) + cot⁻¹(1/x). Let me compute tan⁻¹(x) ≈ tan⁻¹(-5.828) ≈ -1.42 radians. Then cot⁻¹(1/x) = cot⁻¹(-1/5.828) = cot⁻¹(-0.172). As before, cot⁻¹(-0.172) = π - cot⁻¹(0.172) ≈ π - 0.17 ≈ 2.97 radians. Therefore, adding tan⁻¹(x) + cot⁻¹(1/x) ≈ -1.42 + 2.97 ≈ 1.55 radians. Then sin(1.55) ≈ 0.999. But according to the equation, it should be 1/3. Therefore, something is wrong here. But earlier, we had that -2x/(1 + x²) = 1/3. Let me check that for x = -3 - 2√2. As we did before, it gives 1/3. Therefore, according to our previous derivation, sin(π - 2 tan⁻¹(-x)) = 1/3. But π - 2 tan⁻¹(-x) is equal to π - 2 tan⁻¹(-x). Let me compute tan⁻¹(-x) where x = -3 - 2√2. So -x = 3 + 2√2. Therefore, tan⁻¹(-x) = tan⁻¹(3 + 2√2). Let me compute tan⁻¹(3 + 2√2). Let me note that tan(π/8) = √2 - 1 ≈ 0.414, tan(3π/8) = √2 + 1 ≈ 2.414. Wait, 3 + 2√2 ≈ 3 + 2.828 ≈ 5.828. Let me compute tan(π/2 - something). Wait, tan(π/2 - θ) = cot θ. Let me think. Alternatively, note that tan(π/4 + θ) = (1 + tan θ)/(1 - tan θ). Maybe there's an identity here. Alternatively, let me compute tan(π/2 - arctan(3 + 2√2))? Not sure. Alternatively, let me compute tan(π/2 - α) = cot α. If I can find α such that tan α = 3 + 2√2. Let me compute tan(π/8) = √2 - 1 ≈ 0.414, tan(3π/8) = √2 + 1 ≈ 2.414. tan(5π/8) = tan(π - 3π/8) = -tan(3π/8) ≈ -2.414. tan(7π/8) = tan(π - π/8) = -tan(π/8) ≈ -0.414. So tan(π/2 - α) = cot α. If I can find α such that tan α = 3 + 2√2. Let me compute tan(π/2 - α) = cot α = 3 + 2√2. Therefore, π/2 - α = arccot(3 + 2√2). But I don't know the exact value. However, let me note that 3 + 2√2 is equal to (1 + √2)^2. Let me check: (1 + √2)^2 = 1 + 2√2 + 2 = 3 + 2√2. Yes! Therefore, tan α = (1 + √2)^2. Let me think about tan(π/8) and tan(3π/8). Wait, tan(π/8) = √2 - 1 ≈ 0.414, tan(3π/8) = √2 + 1 ≈ 2.414. Let me compute tan(π/4 + π/8) = tan(3π/8) = √2 + 1. Similarly, tan(π/4 + 3π/8) = tan(5π/8) = -tan(3π/8) = -√2 - 1. Not sure. Alternatively, let me consider that tan(π/2 - α) = cot α = 3 + 2√2. Therefore, α = arccot(3 + 2√2). But cot α = 3 + 2√2, so tan α = 1/(3 + 2√2). Let me rationalize 1/(3 + 2√2) = (3 - 2√2)/[(3 + 2√2)(3 - 2√2)] = (3 - 2√2)/(9 - 8) = 3 - 2√2. Therefore, tan α = 3 - 2√2. Therefore, α = arctan(3 - 2√2). Let me compute 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172. Therefore, α ≈ arctan(0.172) ≈ 0.17 radians. Therefore, π/2 - α ≈ π/2 - 0.17 ≈ 1.47 radians. Therefore, tan(π/2 - α) = cot α = 3 + 2√2. Therefore, π/2 - α = arctan(3 + 2√2). Therefore, arctan(3 + 2√2) = π/2 - arctan(3 - 2√2). Therefore, going back, tan⁻¹(-x) = tan⁻¹(3 + 2√2) = π/2 - arctan(3 - 2√2). Therefore, 2 tan⁻¹(-x) = π - 2 arctan(3 - 2√2). Therefore, π - 2 tan⁻¹(-x) = 2 arctan(3 - 2√2). Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2 arctan(3 - 2√2)). Let me compute sin(2 arctan(3 - 2√2)). Let me denote β = arctan(3 - 2√2). Then tan β = 3 - 2√2. Then sin(2β) = 2 tan β / (1 + tan² β). Let me compute tan β = 3 - 2√2. Then tan² β = (3 - 2√2)^2 = 9 - 12√2 + 8 = 17 - 12√2. Therefore, 1 + tan² β = 18 - 12√2. Therefore, sin(2β) = 2*(3 - 2√2)/(18 - 12√2). Let me compute this:

Multiply numerator and denominator by the conjugate of the denominator's denominator? Wait, let me rationalize:

Numerator: 2*(3 - 2√2) = 6 - 4√2.

Denominator: 18 - 12√2 = 6*(3 - 2√2).

Therefore, sin(2β) = (6 - 4√2)/[6*(3 - 2√2)] = [2*(3 - 2√2)]/[6*(3 - 2√2)] = 2/6 = 1/3. Therefore, sin(2β) = 1/3. Therefore, sin(π - 2 tan⁻¹(-x)) = 1/3. Therefore, this is correct. Therefore, even though numerically I thought that sin(1.55) ≈ 0.999, which conflicts, but actually, the angle π - 2 tan⁻¹(-x) is equal to 2β, which is 2 arctan(3 - 2√2). Let me compute 2β. Since β = arctan(3 - 2√2) ≈ arctan(0.172) ≈ 0.17 radians. Therefore, 2β ≈ 0.34 radians. Then sin(0.34) ≈ 0.333, which is 1/3. Therefore, my previous numerical calculation was wrong because I miscalculated the angle. Earlier, I thought that π - 2 tan⁻¹(-x) ≈ 1.55 radians, but actually, π - 2 tan⁻¹(-x) = 2β ≈ 0.34 radians. Wait, how?

Wait, earlier I said that tan⁻¹(-x) = π/2 - β, where β = arctan(3 - 2√2). Therefore, 2 tan⁻¹(-x) = π - 2β. Therefore, π - 2 tan⁻¹(-x) = 2β. Therefore, sin(π - 2 tan⁻¹(-x)) = sin(2β) = 1/3. Therefore, the angle π - 2 tan⁻¹(-x) is 2β ≈ 0.34 radians, not 1.55 radians. Therefore, my previous mistake was in calculating π - 2 tan⁻¹(-x). Let me recalculate:

For x = -3 - 2√2 ≈ -5.828, tan⁻¹(-x) = tan⁻¹(3 + 2√2) ≈ π/2 - β ≈ π/2 - 0.17 ≈ 1.47 radians. Therefore, 2 tan⁻¹(-x) ≈ 2.94 radians. Therefore, π - 2 tan⁻¹(-x) ≈ π - 2.94 ≈ 0.2 radians. Wait, π ≈ 3.14, so 3.14 - 2.94 ≈ 0.2 radians. Therefore, sin(0.2) ≈ 0.198, which is not 1/3. Wait, this is conflicting with the previous result. What is happening here?

Wait, there must be confusion in the angle calculations. Let me try again. Let me denote β = arctan(3 - 2√2). Then tan β = 3 - 2√2. Then, as we saw, sin(2β) = 1/3. Therefore, 2β = arcsin(1/3) or π - arcsin(1/3). Let me compute arcsin(1/3) ≈ 0.3398 radians. Therefore, 2β ≈ 0.3398 radians, so β ≈ 0.1699 radians. Therefore, tan β ≈ tan(0.1699) ≈ 0.172, which matches 3 - 2√2 ≈ 0.172. Therefore, β ≈ 0.17 radians. Therefore, tan⁻¹(-x) = π/2 - β ≈ π/2 - 0.17 ≈ 1.57 - 0.17 ≈ 1.4 radians. Therefore, 2 tan⁻¹(-x) ≈ 2.8 radians. Therefore, π - 2 tan⁻¹(-x) ≈ π - 2.8 ≈ 0.34 radians. Therefore, sin(π - 2 tan⁻¹(-x)) = sin(0.34) ≈ 0.333, which is 1/3. Therefore, the angle π - 2 tan⁻¹(-x) is approximately 0.34 radians, not 1.55 radians. My previous mistake was in calculating π - 2 tan⁻¹(-x). I incorrectly thought that tan⁻¹(-x) was -1.42 radians, but actually tan⁻¹(-x) = tan⁻¹(3 + 2√2) ≈ π/2 - β ≈ 1.4 radians, not -1.42 radians. Wait, no. Wait, x = -3 - 2√2, so -x = 3 + 2√2. Therefore, tan⁻¹(-x) = tan⁻¹(3 + 2√2). Since 3 + 2√2 is positive, tan⁻¹(3 + 2√2) is in (0, π/2). Wait, no! Wait, tan⁻¹(y) for y > 0 is in (0, π/2). Therefore, tan⁻¹(3 + 2√2) is in (0, π/2). Therefore, tan⁻¹(-x) is in (0, π/2). Therefore, π - 2 tan⁻¹(-x) is π - 2*(something in (0, π/2)). Therefore, π - 2*(something in (0, π/2)) is in (π - π, π - 0) = (0, π). Therefore, sin(π - 2 tan⁻¹(-x)) is in (0, 1]. Therefore, sin(π - 2 tan⁻¹(-x)) = 1/3 is valid. Therefore, my previous calculation where I thought tan⁻¹(x) ≈ -1.42 radians was wrong. Wait, tan⁻¹(x) where x = -3 - 2√2 is tan⁻¹(-5.828), which is in (-π/2, 0). Therefore, tan⁻¹(x) ≈ -1.42 radians. But cot⁻¹(1/x) = cot⁻¹(-1/5.828) = π - cot⁻¹(1/5.828) ≈ π - tan⁻¹(5.828) ≈ π - 1.42 ≈ 1.72 radians. Therefore, adding tan⁻¹(x) + cot⁻¹(1/x) ≈ -1.42 + 1.72 ≈ 0.3 radians. Therefore, sin(0.3) ≈ 0.295 ≈ 1/3. Therefore, this is correct. Therefore, my previous mistake was in calculating π - 2 tan⁻¹(-x) as 0.34 radians, which is the same as tan⁻¹(x) + cot⁻¹(1/x) ≈ 0.3 radians. Therefore, everything checks out. Therefore, all four solutions are valid.

Therefore, the solutions are x = 3 + 2√2, x = 3 - 2√2, x = -3 + 2√2, and x = -3 - 2√2.

But let me check if there are any other solutions. For example, when solving the equation sin(θ) = 1/3, θ can be in multiple quadrants. However, in our case, θ = tan⁻¹(x) + cot⁻¹(1/x). For x > 0, θ = 2 tan⁻¹(x), which is between 0 and π. For x < 0, θ = π - 2 tan⁻¹(-x), which is between 0 and π. Therefore, sin(θ) = 1/3 has two solutions in (0, π): arcsin(1/3) and π - arcsin(1/3). Therefore, for each case (x > 0 and x < 0), we have two solutions. Therefore, total four solutions. Therefore, our four solutions are correct.

Therefore, the final answer is all four solutions: 3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

But let me check if these can be simplified or written differently. For example, 3 - 2√2 is already simplified, and -3 + 2√2 is also simplified. Similarly for the others. Therefore, the solutions are:

x = 3 + 2√2,

x = 3 - 2√2,

x = -3 + 2√2,

x = -3 - 2√2.

Therefore, the answer should be these four values.

But let me check if there's any mistake in the quadratic equations. For x > 0, we had 2x/(1 + x²) = 1/3 leading to x² - 6x + 1 = 0, solutions 3 ± 2√2. For x < 0, we had -2x/(1 + x²) = 1/3 leading to x² + 6x + 1 = 0, solutions -3 ± 2√2. Therefore, yes, that's correct. Therefore, the four solutions are correct.

Therefore, the solutions are:

3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

But let me check if these are all distinct. Yes, as we saw numerically, they are four distinct real numbers.

Therefore, the final answer is these four values.

**Final Answer**
The solutions are $\boxed{-3 - 2\sqrt{2}}$, $\boxed{-3 + 2\sqrt{2}}$, $\boxed{3 - 2\sqrt{2}}$, and $\boxed{3 + 2\sqrt{2}}$.

To solve the equation:

$$
\sin \left( \tan^{-1} (x) + \cot^{-1} \left( \frac{1}{x} \right) \right) = \frac{1}{3},
$$

we begin by analyzing the expression inside the sine function.

---

### Step 1: Simplify the Argument of the Sine Function

We use the identity:

$$
\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right),
$$

which holds for $ y > 0 $. Therefore, for $ x > 0 $, we have:

$$
\cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(x).
$$

So the expression becomes:

$$
\tan^{-1}(x) + \cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(x) + \tan^{-1}(x) = 2 \tan^{-1}(x).
$$

Thus, the equation becomes:

$$
\sin\left(2 \tan^{-1}(x)\right) = \frac{1}{3}.
$$

---

### Step 2: Use the Double Angle Identity

Recall the identity:

$$
\sin(2\theta) = 2 \sin(\theta) \cos(\theta).
$$

Let $ \theta = \tan^{-1}(x) $. Then:

- $ \sin(\theta) = \frac{x}{\sqrt{1 + x^2}} $,
- $ \cos(\theta) = \frac{1}{\sqrt{1 + x^2}} $.

So:

$$
\sin(2\theta) = 2 \cdot \frac{x}{\sqrt{1 + x^2}} \cdot \frac{1}{\sqrt{1 + x^2}} = \frac{2x}{1 + x^2}.
$$

Therefore, the equation becomes:

$$
\frac{2x}{1 + x^2} = \frac{1}{3}.
$$

---

### Step 3: Solve the Equation

Cross-multiplying:

$$
6x = 1 + x^2 \quad \Rightarrow \quad x^2 - 6x + 1 = 0.
$$

Solving this quadratic:

$$
x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}.
$$

So, the two positive solutions are:

$$
x = 3 + 2\sqrt{2}, \quad x = 3 - 2\sqrt{2}.
$$

---

### Step 4: Consider Negative Values of $ x $

Now consider $ x < 0 $. In this case, the identity $ \cot^{-1}(1/x) = \tan^{-1}(x) $ does **not** hold, because $ \cot^{-1}(1/x) $ for negative $ x $ lies in the second quadrant, while $ \tan^{-1}(x) $ lies in the fourth.

Instead, we use the identity:

$$
\tan^{-1}(x) + \cot^{-1}\left(\frac{1}{x}\right) = \pi - 2 \tan^{-1}(-x),
$$

for $ x < 0 $. Then:

$$
\sin\left(\pi - 2 \tan^{-1}(-x)\right) = \sin\left(2 \tan^{-1}(-x)\right) = \frac{1}{3}.
$$

Let $ y = -x $, so $ y > 0 $. Then:

$$
\sin\left(2 \tan^{-1}(y)\right) = \frac{1}{3} \quad \Rightarrow \quad \frac{2y}{1 + y^2} = \frac{1}{3}.
$$

This is the same equation as before, leading to:

$$
y^2 - 6y + 1 = 0 \quad \Rightarrow \quad y = 3 \pm 2\sqrt{2}.
$$

Since $ y = -x $, we get:

$$
x = -3 \pm 2\sqrt{2}.
$$

---

### Step 5: Final Answer

Combining all four solutions, we have:

$$
x = 3 + 2\sqrt{2}, \quad x = 3 - 2\sqrt{2}, \quad x = -3 + 2\sqrt{2}, \quad x = -3 - 2\sqrt{2}.
$$

All four values are valid and distinct.

---

### Final Answer

$$
\boxed{-3 - 2\sqrt{2}}, \quad \boxed{-3 + 2\sqrt{2}}, \quad \boxed{3 - 2\sqrt{2}}, \quad \boxed{3 + 2\sqrt{2}}
$$

Solution:
Since $\cot^{-1} \left( \frac{1}{x} \right) = \tan^{-1} x$ for all $x,$ we can write
\[\sin \left( 2 \tan^{-1} x \right) = \frac{1}{3}.\]Let $\theta = \tan^{-1} x,$ so $x = \tan \theta.$  Also, $\sin 2 \theta = \frac{1}{3},$ so
\[2 \sin \theta \cos \theta = \frac{1}{3}.\]Construct a right triangle with legs 1 and $x.$  Then the angle opposite the side length $x$ be $\theta.$

[asy]
unitsize(1 cm);

pair A, B, C;

A = (2,1.8);
B = (0,0);
C = (2,0);

draw(A--B--C--cycle);
draw(rightanglemark(A,C,B,8));

label("$\theta$", B + (0.7,0.3));
label("$1$", (B + C)/2, S);
label("$x$", (A + C)/2, E);
label("$\sqrt{x^2 + 1}$", (A + B)/2, NW);
[/asy]

Also, the hypotenuse will be $\sqrt{x^2 + 1},$ so $\cos \theta = \frac{1}{\sqrt{x^2 + 1}}$ and $\sin \theta = \frac{x}{\sqrt{x^2 + 1}}.$  Hence,
\[2 \cdot \frac{1}{\sqrt{x^2 + 1}} \cdot \frac{x}{\sqrt{x^2 + 1}} = \frac{1}{3},\]or
\[\frac{2x}{x^2 + 1} = \frac{1}{3}.\]This gives us $x^2 + 1 = 6x,$ or $x^2 - 6x + 1 = 0.$  By the quadratic formula, the roots are $x = \boxed{3 \pm 2 \sqrt{2}}.$