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 regions X, Y, and Z. The diagram shows three concentric circles with radii 4, 6, and 7. Let me try to visualize this.

First, concentric circles mean they all share the same center. The largest circle has a radius of 7, then the middle one is 6, and the smallest is 4. The regions X, Y, Z are labeled along the horizontal axis. From the Asymptote code, it seems like X is near the outer edge, Y is between the middle and outer circles, and Z is between the smallest and middle circles? Wait, let me check the Asymptote code again.

The Asymptote code labels X at (2,0), Y at (5,0), and Z at (6.5,0). So, if the circles are centered at (0,0), then X is at (2,0), which is inside the middle circle (radius 6) but outside the smallest circle (radius 4). Wait, radius 4 is smaller than 6, so the region between radius 4 and 6 would be the middle annulus. But then Y is at (5,0), which is still within the middle circle (radius 6) but outside the smallest circle (radius 4). Wait, but maybe the regions are different? Let me think.

The Asymptote code draws the largest circle (radius 7) as light gray, the middle circle (radius 6) as gray, and the smallest circle (radius 4) as white. So, the regions would be:

- The area inside the smallest circle (radius 4) is white.

- The area between radius 4 and 6 is gray (since the middle circle is filled gray, but the inner circle is white, so the annulus between 4 and 6 is gray? Wait, no. Wait, the Asymptote code says:

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

So the largest circle is filled light gray. Then filldraw(circle((0,0),6), gray, black+linewidth(1)); so the middle circle is filled gray, which would cover the inner part of the light gray circle. Then filldraw(circle((0,0),4), white, black+linewidth(1)); so the smallest circle is filled white, covering the inner part of the gray circle.

Therefore, the regions would be:

- The innermost region (radius 4) is white.

- The region between radius 4 and 6 is gray (since the middle circle is filled gray, but the inner circle is white, so the annulus between 4 and 6 is gray).

- The region between radius 6 and 7 is light gray (since the largest circle is filled light gray, and the middle circle is filled gray, which is darker, so the annulus between 6 and 7 is light gray).

But the problem says three regions are labeled X, Y, Z. The labels are at (2,0), (5,0), (6.5,0). So:

- X is at (2,0): this is in the annulus between 4 and 6? Wait, (2,0) is at radius 2, which is inside the smallest circle (radius 4). Wait, but the innermost region is white. Wait, but the label X is at (2,0), which is inside the smallest circle. But the Asymptote code labels X at (2,0), which is inside the smallest circle? But the smallest circle is white. However, the problem says three regions are labeled X, Y, Z. Maybe the regions are:

- The innermost circle (radius 4) is white, labeled Z?

Wait, maybe I need to think again. The Asymptote code draws the largest circle (radius 7) as light gray, then the middle circle (radius 6) as gray, which would cover the inner part of the light gray, making the area between 6 and 7 light gray, and the area between 4 and 6 gray, and the innermost circle (radius 4) white. So the regions are:

- White region: inside radius 4.

- Gray region: between 4 and 6.

- Light gray region: between 6 and 7.

But the labels X, Y, Z are placed at (2,0), (5,0), (6.5,0). So:

- X is at (2,0): inside radius 4, so white region.

- Y is at (5,0): between 4 and 6, so gray region.

- Z is at (6.5,0): between 6 and 7, so light gray region.

But the problem says "three regions are labeled X, Y, or Z below." So maybe the regions are:

- X: white (innermost circle)

- Y: gray (annulus between 4 and 6)

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

But 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 we need to compute the areas of X, Y, Z and find the difference between the largest and smallest.

But let me confirm. The Asymptote code labels X at (2,0), which is inside the smallest circle (radius 4). So that's the white region. Y is at (5,0), which is between 4 and 6, so the gray region. Z is at (6.5,0), which is between 6 and 7, so the light gray region. Therefore, the three regions are:

- X: area of the innermost circle (radius 4)

- Y: area of the annulus between 4 and 6

- Z: area of the annulus between 6 and 7

Therefore, compute each area:

First, area of X: π*(4)^2 = 16π

Area of Y: area between 4 and 6: π*(6)^2 - π*(4)^2 = 36π - 16π = 20π

Area of Z: area between 6 and 7: π*(7)^2 - π*(6)^2 = 49π - 36π = 13π

Wait, so then the areas are:

X: 16π

Y: 20π

Z: 13π

Therefore, the largest area is Y (20π), the smallest is Z (13π). The difference is 20π - 13π = 7π. But wait, hold on. Wait, X is 16π, Y is 20π, Z is 13π. So the largest is Y (20π), the smallest is Z (13π). So difference is 7π. But let me check again.

Wait, but maybe I got the regions wrong. Let me check again. The Asymptote code labels X at (2,0), which is inside the smallest circle (radius 4). But the problem says "three regions are labeled X, Y, or Z below." Maybe the regions are not the annuli but something else? Wait, the Asymptote code draws three circles, with the innermost white, middle gray, outer light gray. So the regions are:

- The white circle (radius 4)

- The gray annulus (between 4 and 6)

- The light gray annulus (between 6 and 7)

Therefore, the three regions are X, Y, Z. The labels are placed at (2,0), (5,0), (6.5,0). So X is in the white region, Y in the gray, Z in the light gray. Therefore, their areas are as calculated above. So difference between largest (Y: 20π) and smallest (Z: 13π) is 7π. But let me check again.

Wait, but maybe the regions are different? Maybe X is the annulus between 4 and 6? Wait, the label X is at (2,0), which is inside the smallest circle. So if the regions are labeled X, Y, Z, maybe X is the innermost circle, Y is the middle annulus, Z is the outer annulus. So then the areas are as above. So difference is 20π - 13π = 7π. But let me check again.

Alternatively, maybe the regions are different. Maybe the regions are the areas between the circles? Wait, but there are three circles, so there are two annuli and the inner circle. So three regions. So the problem says "three regions are labeled X, Y, or Z below." So the answer would be 7π. But let me check again.

Alternatively, maybe the regions are not the annuli but something else? Wait, the Asymptote code draws the largest circle as light gray, then the middle circle as gray, which would cover the inner part of the light gray, and then the smallest circle as white, covering the inner part of the gray. Therefore, the regions are:

- The innermost circle (white)

- The middle annulus (gray)

- The outer annulus (light gray)

Therefore, the three regions. So their areas are:

White: π*(4)^2 = 16π

Gray: π*(6)^2 - π*(4)^2 = 36π - 16π = 20π

Light gray: π*(7)^2 - π*(6)^2 = 49π - 36π = 13π

So the areas are 16π, 20π, 13π. Therefore, the largest area is 20π (Y), the smallest is 13π (Z). The difference is 20π - 13π = 7π. Therefore, the answer is 7π. But let me check once more.

Wait, but maybe the regions are different? For example, maybe X is the outer annulus, Y is the middle annulus, Z is the inner circle? But according to the Asymptote code, the labels are at (2,0), (5,0), (6.5,0). So (2,0) is near the center, so that would be the inner circle. (5,0) is between 4 and 6, so middle annulus. (6.5,0) is between 6 and 7, outer annulus. Therefore, X is inner circle, Y is middle annulus, Z is outer annulus. Therefore, areas as above. So difference between largest (Y: 20π) and smallest (Z: 13π) is 7π. But wait, the inner circle is 16π, which is bigger than Z's 13π. Wait, so the smallest area is Z (13π), and the largest is Y (20π). Therefore, difference is 7π. But let me check again.

Wait, 16π is bigger than 13π. So the regions have areas:

X: 16π

Y: 20π

Z: 13π

So the largest is Y (20π), the smallest is Z (13π). Therefore, difference is 20π - 13π = 7π. So answer is 7π. But let me check once more.

Alternatively, maybe the regions are different. Maybe X is the outer annulus, Y is the middle annulus, Z is the inner circle? But according to the labels, X is at (2,0), which is near the center, so that would be the inner circle. Y is at (5,0), which is between 4 and 6, so middle annulus. Z is at (6.5,0), which is between 6 and 7, outer annulus. Therefore, the areas are as above. So the answer is 7π.

But let me check if I made a mistake in calculating the areas.

Area of inner circle (X): π*(4)^2 = 16π. Correct.

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

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

So the areas are 16π, 20π, 13π. Therefore, the difference between the largest (20π) and smallest (13π) is 7π. Therefore, the answer is 7π.

But 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 greatest area is Y (20π) and the smallest is Z (13π). So difference is 7π.

But let me check once more if the regions are labeled correctly. The Asymptote code labels X at (2,0), which is inside the smallest circle (radius 4). So that's the inner circle. Y is at (5,0), which is between 4 and 6, so middle annulus. Z is at (6.5,0), which is between 6 and 7, outer annulus. Therefore, the areas are as calculated.

Alternatively, maybe the problem is in the interpretation of the regions. Maybe the regions are not the annuli but something else? For example, maybe X is the area between 4 and 6, Y between 6 and 7, and Z the inner circle? But according to the Asymptote code, the labels are placed at (2,0), (5,0), (6.5,0). So (2,0) is inside the inner circle, so that would be Z? Wait, but the problem says "three regions are labeled X, Y, or Z below." Maybe the labels are not corresponding to the regions? Wait, the Asymptote code is:

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 light gray, then the middle circle is gray, which covers the inner part of the light gray, and the smallest circle is white, covering the inner part of the gray. Therefore, the regions are:

- The innermost circle (white)

- The middle annulus (gray)

- The outer annulus (light gray)

Therefore, the three regions. The labels are placed at (2,0), (5,0), (6.5,0). So:

- (2,0) is in the innermost circle (white), so that's region X.

- (5,0) is in the middle annulus (gray), so that's region Y.

- (6.5,0) is in the outer annulus (light gray), so that's region Z.

Therefore, the areas are as calculated. Therefore, the answer is 7π.

But let me check once more. Maybe the problem is in the Asymptote code? The Asymptote code draws the circles with filldraw, so the innermost circle is white, then the middle circle is gray, which would cover the inner part of the light gray, so the middle annulus is gray, and the outer annulus is light gray. Therefore, the regions are:

- White: inner circle (radius 4)

- Gray: annulus between 4 and 6

- Light gray: annulus between 6 and 7

Therefore, the areas are:

White: π*(4)^2 = 16π

Gray: π*(6)^2 - π*(4)^2 = 20π

Light gray: π*(7)^2 - π*(6)^2 = 13π

So the regions X, Y, Z have areas 16π, 20π, 13π. Therefore, the largest area is 20π (Y), the smallest is 13π (Z). The difference is 20π - 13π = 7π.

Therefore, the answer is 7π.

But let me check if there's any chance that the regions are different. For example, maybe X is the outer annulus? But according to the Asymptote code, the label X is at (2,0), which is near the center, so it's more likely to be the inner circle.

Alternatively, maybe the problem is in the order of the regions? For example, maybe the regions are labeled differently? But the Asymptote code labels X at (2,0), Y at (5,0), Z at (6.5,0). So X is closest to the center, then Y, then Z. Therefore, the areas would be increasing from X to Y to Z? Wait, no. Wait, the inner circle is 16π, middle annulus is 20π, outer annulus is 13π. So the areas are X: 16π, Y: 20π, Z: 13π. Therefore, the order from smallest to largest is Z (13π), X (16π), Y (20π). Therefore, the difference between the largest (Y) and smallest (Z) is 7π.

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

But let me check once more. Maybe I made a mistake in the area calculations.

Area of the inner circle (radius 4): πr² = π*(4)^2 = 16π. Correct.

Area of the middle annulus (between 4 and 6): π*(6² - 4²) = π*(36 - 16) = 20π. Correct.

Area of the outer annulus (between 6 and 7): π*(7² - 6²) = π*(49 - 36) = 13π. Correct.

Yes, that's correct. Therefore, the difference is 20π - 13π = 7π.

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 $, and we are asked to find the difference between the area of the region with the greatest area and the area of the region with the smallest area.

---

### Step 1: Understand the Regions

From the Asymptote code and the diagram:

- The **innermost circle** (radius 4) is **white**.
- The **middle annulus** (between radii 4 and 6) is **gray**.
- The **outer annulus** (between radii 6 and 7) is **light gray**.

The labels are placed at:

- $ X $ at (2,0): inside the innermost circle → **white region**
- $ Y $ at (5,0): between 4 and 6 → **gray annulus**
- $ Z $ at (6.5,0): between 6 and 7 → **light gray annulus**

So, the three regions are:

- **Region X**: Inner circle (radius 4)
- **Region Y**: Annulus between 4 and 6
- **Region Z**: Annulus between 6 and 7

---

### Step 2: Calculate the Areas

We compute the area of each region using the formula for the area of a circle, $ A = \pi r^2 $, and the area of an annulus, $ A = \pi (R^2 - r^2) $, where $ R $ is the outer radius and $ r $ is the inner radius.

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

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

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

---

### Step 3: Identify the Largest and Smallest Areas

From the computed areas:

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

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

---

### Step 4: 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 regions X, Y, and Z. The diagram shows three concentric circles with radii 4, 6, and 7. Let me try to visualize this.

First, concentric circles mean they all share the same center. The largest circle has a radius of 7, then the middle one is 6, and the smallest is 4. The regions X, Y, Z are labeled along the horizontal axis. From the Asymptote code, it seems like X is near the outer edge, Y is between the middle and outer circles, and Z is between the smallest and middle circles? Wait, let me check the Asymptote code again.

The Asymptote code labels X at (2,0), Y at (5,0), and Z at (6.5,0). So, if the circles are centered at (0,0), then X is at (2,0), which is inside the middle circle (radius 6) but outside the smallest circle (radius 4). Wait, radius 4 is smaller than 6, so the region between radius 4 and 6 would be the middle annulus. But then Y is at (5,0), which is still within the middle circle (radius 6) but outside the smallest circle (radius 4). Wait, but maybe the regions are different? Let me think.

The Asymptote code draws the largest circle (radius 7) as light gray, the middle circle (radius 6) as gray, and the smallest circle (radius 4) as white. So, the regions would be:

- The area inside the smallest circle (radius 4) is white.

- The area between radius 4 and 6 is gray (since the middle circle is filled gray, but the inner circle is white, so the annulus between 4 and 6 is gray? Wait, no. Wait, the Asymptote code says:

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

So the largest circle is filled light gray. Then filldraw(circle((0,0),6), gray, black+linewidth(1)); so the middle circle is filled gray, which would cover the inner part of the light gray circle. Then filldraw(circle((0,0),4), white, black+linewidth(1)); so the smallest circle is filled white, covering the inner part of the gray circle.

Therefore, the regions would be:

- The innermost region (radius 4) is white.

- The region between radius 4 and 6 is gray (since the middle circle is filled gray, but the inner circle is white, so the annulus between 4 and 6 is gray).

- The region between radius 6 and 7 is light gray (since the largest circle is filled light gray, and the middle circle is filled gray, which is darker, so the annulus between 6 and 7 is light gray).

But the problem says three regions are labeled X, Y, Z. The labels are at (2,0), (5,0), (6.5,0). So:

- X is at (2,0): this is in the annulus between 4 and 6? Wait, (2,0) is at radius 2, which is inside the smallest circle (radius 4). Wait, but the innermost region is white. Wait, but the label X is at (2,0), which is inside the smallest circle. But the Asymptote code labels X at (2,0), which is inside the smallest circle? But the smallest circle is white. However, the problem says three regions are labeled X, Y, Z. Maybe the regions are:

- The innermost circle (radius 4) is white, labeled Z?

Wait, maybe I need to think again. The Asymptote code draws the largest circle (radius 7) as light gray, then the middle circle (radius 6) as gray, which would cover the inner part of the light gray, making the area between 6 and 7 light gray, and the area between 4 and 6 gray, and the innermost circle (radius 4) white. So the regions are:

- White region: inside radius 4.

- Gray region: between 4 and 6.

- Light gray region: between 6 and 7.

But the labels X, Y, Z are placed at (2,0), (5,0), (6.5,0). So:

- X is at (2,0): inside radius 4, so white region.

- Y is at (5,0): between 4 and 6, so gray region.

- Z is at (6.5,0): between 6 and 7, so light gray region.

But the problem says "three regions are labeled X, Y, or Z below." So maybe the regions are:

- X: white (innermost circle)

- Y: gray (annulus between 4 and 6)

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

But 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 we need to compute the areas of X, Y, Z and find the difference between the largest and smallest.

But let me confirm. The Asymptote code labels X at (2,0), which is inside the smallest circle (radius 4). So that's the white region. Y is at (5,0), which is between 4 and 6, so the gray region. Z is at (6.5,0), which is between 6 and 7, so the light gray region. Therefore, the three regions are:

- X: area of the innermost circle (radius 4)

- Y: area of the annulus between 4 and 6

- Z: area of the annulus between 6 and 7

Therefore, compute each area:

First, area of X: π*(4)^2 = 16π

Area of Y: area between 4 and 6: π*(6)^2 - π*(4)^2 = 36π - 16π = 20π

Area of Z: area between 6 and 7: π*(7)^2 - π*(6)^2 = 49π - 36π = 13π

Wait, so then the areas are:

X: 16π

Y: 20π

Z: 13π

Therefore, the largest area is Y (20π), the smallest is Z (13π). The difference is 20π - 13π = 7π. But wait, hold on. Wait, X is 16π, Y is 20π, Z is 13π. So the largest is Y (20π), the smallest is Z (13π). So difference is 7π. But let me check again.

Wait, but maybe I got the regions wrong. Let me check again. The Asymptote code labels X at (2,0), which is inside the smallest circle (radius 4). But the problem says "three regions are labeled X, Y, or Z below." Maybe the regions are not the annuli but something else? Wait, the Asymptote code draws three circles, with the innermost white, middle gray, outer light gray. So the regions are:

- The white circle (radius 4)

- The gray annulus (between 4 and 6)

- The light gray annulus (between 6 and 7)

Therefore, the three regions are X, Y, Z. The labels are placed at (2,0), (5,0), (6.5,0). So X is in the white region, Y in the gray, Z in the light gray. Therefore, their areas are as calculated above. So difference between largest (Y: 20π) and smallest (Z: 13π) is 7π. But let me check again.

Wait, but maybe the regions are different? Maybe X is the annulus between 4 and 6? Wait, the label X is at (2,0), which is inside the smallest circle. So if the regions are labeled X, Y, Z, maybe X is the innermost circle, Y is the middle annulus, Z is the outer annulus. So then the areas are as above. So difference is 20π - 13π = 7π. But let me check again.

Alternatively, maybe the regions are different. Maybe the regions are the areas between the circles? Wait, but there are three circles, so there are two annuli and the inner circle. So three regions. So the problem says "three regions are labeled X, Y, or Z below." So the answer would be 7π. But let me check again.

Alternatively, maybe the regions are not the annuli but something else? Wait, the Asymptote code draws the largest circle as light gray, then the middle circle as gray, which would cover the inner part of the light gray, and then the smallest circle as white, covering the inner part of the gray. Therefore, the regions are:

- The innermost circle (white)

- The middle annulus (gray)

- The outer annulus (light gray)

Therefore, the three regions. So their areas are:

White: π*(4)^2 = 16π

Gray: π*(6)^2 - π*(4)^2 = 36π - 16π = 20π

Light gray: π*(7)^2 - π*(6)^2 = 49π - 36π = 13π

So the areas are 16π, 20π, 13π. Therefore, the largest area is 20π (Y), the smallest is 13π (Z). The difference is 20π - 13π = 7π. Therefore, the answer is 7π. But let me check once more.

Wait, but maybe the regions are different? For example, maybe X is the outer annulus, Y is the middle annulus, Z is the inner circle? But according to the Asymptote code, the labels are at (2,0), (5,0), (6.5,0). So (2,0) is near the center, so that would be the inner circle. (5,0) is between 4 and 6, so middle annulus. (6.5,0) is between 6 and 7, outer annulus. Therefore, X is inner circle, Y is middle annulus, Z is outer annulus. Therefore, areas as above. So difference between largest (Y: 20π) and smallest (Z: 13π) is 7π. But wait, the inner circle is 16π, which is bigger than Z's 13π. Wait, so the smallest area is Z (13π), and the largest is Y (20π). Therefore, difference is 7π. But let me check again.

Wait, 16π is bigger than 13π. So the regions have areas:

X: 16π

Y: 20π

Z: 13π

So the largest is Y (20π), the smallest is Z (13π). Therefore, difference is 20π - 13π = 7π. So answer is 7π. But let me check once more.

Alternatively, maybe the regions are different. Maybe X is the outer annulus, Y is the middle annulus, Z is the inner circle? But according to the labels, X is at (2,0), which is near the center, so that would be the inner circle. Y is at (5,0), which is between 4 and 6, so middle annulus. Z is at (6.5,0), which is between 6 and 7, outer annulus. Therefore, the areas are as above. So the answer is 7π.

But let me check if I made a mistake in calculating the areas.

Area of inner circle (X): π*(4)^2 = 16π. Correct.

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

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

So the areas are 16π, 20π, 13π. Therefore, the difference between the largest (20π) and smallest (13π) is 7π. Therefore, the answer is 7π.

But 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 greatest area is Y (20π) and the smallest is Z (13π). So difference is 7π.

But let me check once more if the regions are labeled correctly. The Asymptote code labels X at (2,0), which is inside the smallest circle (radius 4). So that's the inner circle. Y is at (5,0), which is between 4 and 6, so middle annulus. Z is at (6.5,0), which is between 6 and 7, outer annulus. Therefore, the areas are as calculated.

Alternatively, maybe the problem is in the interpretation of the regions. Maybe the regions are not the annuli but something else? For example, maybe X is the area between 4 and 6, Y between 6 and 7, and Z the inner circle? But according to the Asymptote code, the labels are placed at (2,0), (5,0), (6.5,0). So (2,0) is inside the inner circle, so that would be Z? Wait, but the problem says "three regions are labeled X, Y, or Z below." Maybe the labels are not corresponding to the regions? Wait, the Asymptote code is:

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 light gray, then the middle circle is gray, which covers the inner part of the light gray, and the smallest circle is white, covering the inner part of the gray. Therefore, the regions are:

- The innermost circle (white)

- The middle annulus (gray)

- The outer annulus (light gray)

Therefore, the three regions. The labels are placed at (2,0), (5,0), (6.5,0). So:

- (2,0) is in the innermost circle (white), so that's region X.

- (5,0) is in the middle annulus (gray), so that's region Y.

- (6.5,0) is in the outer annulus (light gray), so that's region Z.

Therefore, the areas are as calculated. Therefore, the answer is 7π.

But let me check once more. Maybe the problem is in the Asymptote code? The Asymptote code draws the circles with filldraw, so the innermost circle is white, then the middle circle is gray, which would cover the inner part of the light gray, so the middle annulus is gray, and the outer annulus is light gray. Therefore, the regions are:

- White: inner circle (radius 4)

- Gray: annulus between 4 and 6

- Light gray: annulus between 6 and 7

Therefore, the areas are:

White: π*(4)^2 = 16π

Gray: π*(6)^2 - π*(4)^2 = 20π

Light gray: π*(7)^2 - π*(6)^2 = 13π

So the regions X, Y, Z have areas 16π, 20π, 13π. Therefore, the largest area is 20π (Y), the smallest is 13π (Z). The difference is 20π - 13π = 7π.

Therefore, the answer is 7π.

But let me check if there's any chance that the regions are different. For example, maybe X is the outer annulus? But according to the Asymptote code, the label X is at (2,0), which is near the center, so it's more likely to be the inner circle.

Alternatively, maybe the problem is in the order of the regions? For example, maybe the regions are labeled differently? But the Asymptote code labels X at (2,0), Y at (5,0), Z at (6.5,0). So X is closest to the center, then Y, then Z. Therefore, the areas would be increasing from X to Y to Z? Wait, no. Wait, the inner circle is 16π, middle annulus is 20π, outer annulus is 13π. So the areas are X: 16π, Y: 20π, Z: 13π. Therefore, the order from smallest to largest is Z (13π), X (16π), Y (20π). Therefore, the difference between the largest (Y) and smallest (Z) is 7π.

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

But let me check once more. Maybe I made a mistake in the area calculations.

Area of the inner circle (radius 4): πr² = π*(4)^2 = 16π. Correct.

Area of the middle annulus (between 4 and 6): π*(6² - 4²) = π*(36 - 16) = 20π. Correct.

Area of the outer annulus (between 6 and 7): π*(7² - 6²) = π*(49 - 36) = 13π. Correct.

Yes, that's correct. Therefore, the difference is 20π - 13π = 7π.

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 $, and we are asked to find the difference between the area of the region with the greatest area and the area of the region with the smallest area.

---

### Step 1: Understand the Regions

From the Asymptote code and the diagram:

- The **innermost circle** (radius 4) is **white**.
- The **middle annulus** (between radii 4 and 6) is **gray**.
- The **outer annulus** (between radii 6 and 7) is **light gray**.

The labels are placed at:

- $ X $ at (2,0): inside the innermost circle → **white region**
- $ Y $ at (5,0): between 4 and 6 → **gray annulus**
- $ Z $ at (6.5,0): between 6 and 7 → **light gray annulus**

So, the three regions are:

- **Region X**: Inner circle (radius 4)
- **Region Y**: Annulus between 4 and 6
- **Region Z**: Annulus between 6 and 7

---

### Step 2: Calculate the Areas

We compute the area of each region using the formula for the area of a circle, $ A = \pi r^2 $, and the area of an annulus, $ A = \pi (R^2 - r^2) $, where $ R $ is the outer radius and $ r $ is the inner radius.

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

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

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

---

### Step 3: Identify the Largest and Smallest Areas

From the computed areas:

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

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

---

### Step 4: 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. Okay, that seems right.

Alternatively, maybe there's a different identity. Let me think. Another way to approach this is to use the identity for sin(A + B). Let me try that approach as well to confirm.

Let me denote A = tan⁻¹(x) and B = cot⁻¹(1/x). Then, the equation is sin(A + B) = 1/3. Let me compute A + B. If A = tan⁻¹(x), then tan A = x. For B = cot⁻¹(1/x), which is the same as tan⁻¹(x) as we discussed earlier. Therefore, B = tan⁻¹(x). Therefore, A + B = 2 tan⁻¹(x). So again, same result. Therefore, sin(2 tan⁻¹(x)) = 1/3.

Alternatively, maybe there's a relationship between A and B? Let me check. If A = tan⁻¹(x), then cot⁻¹(1/x) is equal to π/2 - tan⁻¹(x)? Wait, no. Wait, tan(π/2 - θ) = cot θ. Therefore, cot⁻¹(y) = π/2 - tan⁻¹(y). Wait, so if we have cot⁻¹(1/x), then that would be π/2 - tan⁻¹(1/x). Wait, but earlier I thought cot⁻¹(1/x) = tan⁻¹(x). Let me check with numbers again. Let me take x = 1. Then cot⁻¹(1/1) = cot⁻¹(1) = π/4, and tan⁻¹(1) is also π/4. So that works. If x = 2, then cot⁻¹(1/2) is the angle whose cotangent is 1/2, which is the same as tan⁻¹(2). Because cot(θ) = 1/tan(θ), so if cot(θ) = 1/2, then tan(θ) = 2, so θ = tan⁻¹(2). Therefore, cot⁻¹(1/x) = tan⁻¹(x). Therefore, that seems correct. Therefore, A + B = tan⁻¹(x) + tan⁻¹(x) = 2 tan⁻¹(x). Therefore, sin(2 tan⁻¹(x)) = 1/3.

Alternatively, if I use the identity for sin(A + B), maybe I can compute sin(A + B) directly. Let me try that. Let me compute sin(A + B) where A = tan⁻¹(x) and B = cot⁻¹(1/x). Let me compute sin(A + B) = sin A cos B + cos A sin B. Let me compute sin A, cos A, sin B, cos B.

Since A = tan⁻¹(x), then we can imagine a right triangle where the opposite side is x and adjacent side is 1, so hypotenuse is √(1 + x²). Therefore, sin A = x / √(1 + x²), cos A = 1 / √(1 + x²).

Similarly, B = cot⁻¹(1/x). Let me think about B. cot B = 1/x. So cot B = adjacent / opposite = 1/x. Therefore, if we imagine a right triangle where adjacent side is 1 and opposite side is x, hypotenuse is √(1 + x²). Therefore, sin B = opposite / hypotenuse = x / √(1 + x²), and cos B = adjacent / hypotenuse = 1 / √(1 + x²).

Therefore, sin(A + B) = sin A cos B + cos A sin B = [x / √(1 + x²) * 1 / √(1 + x²)] + [1 / √(1 + x²) * x / √(1 + x²)] = [x / (1 + x²)] + [x / (1 + x²)] = 2x / (1 + x²). Therefore, sin(A + B) = 2x / (1 + x²). Therefore, the original equation is 2x / (1 + x²) = 1/3.

Wait, so that's another way to get the same equation. Therefore, regardless of the approach, we end up with 2x / (1 + x²) = 1/3. Therefore, solving this equation for x.

So let me write that equation:

2x / (1 + x²) = 1/3.

Cross multiplying:

2x * 3 = 1 * (1 + x²)

Which is:

6x = 1 + x²

Rearranging:

x² - 6x + 1 = 0

Wait, hold on. Wait, 6x = 1 + x² => x² - 6x + 1 = 0? Wait, no. Let me check:

Starting from 6x = 1 + x², subtract 6x from both sides:

0 = x² - 6x + 1, so yes, x² - 6x + 1 = 0. Therefore, solving quadratic equation:

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

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

Wait, but hold on. Let me check if there are any restrictions on x. Since tan⁻¹(x) is defined for all real numbers, and cot⁻¹(1/x) is defined when 1/x is defined, so x ≠ 0. Therefore, x cannot be zero. However, our solutions are 3 + 2√2 and 3 - 2√2. Let me check if they are non-zero. 3 - 2√2 is approximately 3 - 2.828 ≈ 0.172, which is positive, so not zero. Therefore, both solutions are valid.

But wait, let me check if there are any other constraints. For example, when dealing with inverse trigonometric functions, sometimes there can be issues with the domain or range. Let me think. The original equation is sin( tan⁻¹(x) + cot⁻¹(1/x) ) = 1/3. We found that tan⁻¹(x) + cot⁻¹(1/x) simplifies to 2 tan⁻¹(x). However, is there a case where tan⁻¹(x) + cot⁻¹(1/x) is not equal to 2 tan⁻¹(x)? Wait, earlier we saw that cot⁻¹(1/x) = tan⁻¹(x). Let me check for negative x. Suppose x is negative. Let me take x = -1. Then tan⁻¹(-1) = -π/4. cot⁻¹(1/(-1)) = cot⁻¹(-1). Now, cot⁻¹(-1) is equal to π - cot⁻¹(1) = π - π/4 = 3π/4? Wait, but cot⁻¹(y) is usually defined to have a range of (0, π). So cot⁻¹(-1) would be π - cot⁻¹(1) = π - π/4 = 3π/4. However, tan⁻¹(x) for x negative is negative. So in this case, tan⁻¹(-1) + cot⁻¹(-1) = -π/4 + 3π/4 = π/2. However, 2 tan⁻¹(-1) = 2*(-π/4) = -π/2. But sin(-π/2) = -1, whereas sin(π/2) = 1. Therefore, there's a discrepancy here. Therefore, my previous conclusion that cot⁻¹(1/x) = tan⁻¹(x) is only valid for positive x? Wait, so maybe there's a mistake here.

Wait, so if x is negative, then cot⁻¹(1/x) is not equal to tan⁻¹(x). Let me check again. Let me recall that cot⁻¹(y) is equal to tan⁻¹(1/y) when y > 0, but for y < 0, cot⁻¹(y) is equal to tan⁻¹(1/y) + π? Or something like that? Wait, let me recall the exact ranges.

The range of cot⁻¹(y) is typically (0, π). The range of tan⁻¹(y) is (-π/2, π/2). So if we have cot⁻¹(y) = π/2 - tan⁻¹(y) for y > 0. Wait, let me check. For y > 0, cot⁻¹(y) = tan⁻¹(1/y). But tan⁻¹(1/y) is in (-π/2, π/2). However, cot⁻¹(y) is in (0, π). Therefore, for y > 0, cot⁻¹(y) = π/2 - tan⁻¹(y). Wait, let me check with y = 1. cot⁻¹(1) = π/4, and π/2 - tan⁻¹(1) = π/2 - π/4 = π/4. Correct. For y = 2, cot⁻¹(2) = π/2 - tan⁻¹(2). So that's correct. However, for y < 0, cot⁻¹(y) is in (0, π). Let me take y = -1. Then cot⁻¹(-1) is π - cot⁻¹(1) = π - π/4 = 3π/4. On the other hand, tan⁻¹(1/y) = tan⁻¹(-1) = -π/4. Therefore, cot⁻¹(y) = π + tan⁻¹(1/y) for y < 0? Wait, π + tan⁻¹(1/y) would be π + (-π/4) = 3π/4, which matches. Therefore, in general, cot⁻¹(y) = tan⁻¹(1/y) if y > 0, and cot⁻¹(y) = π + tan⁻¹(1/y) if y < 0. Therefore, in general, cot⁻¹(y) = tan⁻¹(1/y) + π * H(y), where H(y) is 0 if y > 0 and 1 if y < 0? Wait, not sure. But in any case, for our problem, we have cot⁻¹(1/x). Let me consider two cases: x > 0 and x < 0.

Case 1: x > 0. Then cot⁻¹(1/x) = tan⁻¹(x). Therefore, tan⁻¹(x) + cot⁻¹(1/x) = tan⁻¹(x) + tan⁻¹(x) = 2 tan⁻¹(x). Therefore, sin(2 tan⁻¹(x)) = 1/3.

Case 2: x < 0. Then cot⁻¹(1/x) = π + tan⁻¹(x). Wait, let me check. If x < 0, then 1/x < 0. Therefore, cot⁻¹(1/x) = π + tan⁻¹(1/(1/x))? Wait, no. Wait, earlier we saw that for y < 0, cot⁻¹(y) = π + tan⁻¹(1/y). Wait, no. Wait, for y < 0, cot⁻¹(y) = π + tan⁻¹(1/y)? Let me check with y = -1. cot⁻¹(-1) = 3π/4. tan⁻¹(1/y) = tan⁻¹(-1) = -π/4. So π + tan⁻¹(1/y) = π - π/4 = 3π/4. Correct. Therefore, for y < 0, cot⁻¹(y) = π + tan⁻¹(1/y). Therefore, in our case, cot⁻¹(1/x) where x < 0, so 1/x < 0. Therefore, cot⁻¹(1/x) = π + tan⁻¹(1/(1/x)) = π + tan⁻¹(x). Therefore, tan⁻¹(x) + cot⁻¹(1/x) = tan⁻¹(x) + π + tan⁻¹(x) = 2 tan⁻¹(x) + π. Therefore, sin(2 tan⁻¹(x) + π) = sin(2 tan⁻¹(x)) * cos(π) + cos(2 tan⁻¹(x)) * sin(π) = -sin(2 tan⁻¹(x)) + 0 = -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, depending on the sign of x, we have different equations. Therefore, my previous approach only considered x > 0, but we need to check both cases.

Therefore, let me re-examine the problem. So, the original equation is:

sin( tan⁻¹(x) + cot⁻¹(1/x) ) = 1/3.

But depending on the sign of x, the expression inside the sine function is either 2 tan⁻¹(x) or 2 tan⁻¹(x) + π. Therefore, we need to consider both cases.

Case 1: x > 0. Then, sin(2 tan⁻¹(x)) = 1/3.

Case 2: x < 0. Then, sin(2 tan⁻¹(x) + π) = 1/3 => -sin(2 tan⁻¹(x)) = 1/3 => sin(2 tan⁻¹(x)) = -1/3.

Therefore, we need to solve both equations:

For x > 0: sin(2 tan⁻¹(x)) = 1/3.

For x < 0: sin(2 tan⁻¹(x)) = -1/3.

But earlier, when I considered the general case, I derived that sin(A + B) = 2x / (1 + x²). Wait, let me check that again. Earlier, I used the identity for sin(A + B) and found that sin(A + B) = 2x / (1 + x²). However, that was under the assumption that A = tan⁻¹(x) and B = cot⁻¹(1/x). But if x is negative, does that formula still hold?

Wait, let me recalculate sin(A + B) for x negative. Let me take x negative. Let me take x = -k where k > 0. Then tan⁻¹(x) = tan⁻¹(-k) = -tan⁻¹(k). Similarly, cot⁻¹(1/x) = cot⁻¹(-1/k). As we saw earlier, cot⁻¹(-1/k) = π + tan⁻¹(-k) = π - tan⁻¹(k). Therefore, A + B = tan⁻¹(x) + cot⁻¹(1/x) = -tan⁻¹(k) + π - tan⁻¹(k) = π - 2 tan⁻¹(k). Therefore, sin(A + B) = sin(π - 2 tan⁻¹(k)) = sin(2 tan⁻¹(k)) because sin(π - θ) = sin θ. Therefore, sin(A + B) = sin(2 tan⁻¹(k)) = 2k / (1 + k²). But since x = -k, then k = -x. Therefore, sin(A + B) = 2*(-x) / (1 + x²). Wait, because k = -x, so 2k / (1 + k²) = 2*(-x) / (1 + x²). Therefore, sin(A + B) = -2x / (1 + x²). Therefore, in this case, sin(A + B) = -2x / (1 + x²). Therefore, the original equation sin(A + B) = 1/3 becomes -2x / (1 + x²) = 1/3. Therefore, for x < 0, we have -2x / (1 + x²) = 1/3. Therefore, combining both cases:

For x > 0: 2x / (1 + x²) = 1/3.

For x < 0: -2x / (1 + x²) = 1/3.

Therefore, solving both equations.

So let me write both equations:

Case 1: 2x / (1 + x²) = 1/3.

Case 2: -2x / (1 + x²) = 1/3 => 2x / (1 + x²) = -1/3.

Therefore, solving both equations.

Starting with Case 1: 2x / (1 + x²) = 1/3.

Multiply both sides by (1 + x²):

2x = (1 + x²)/3 => 6x = 1 + x² => x² - 6x + 1 = 0.

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

Now, check if these solutions are positive. 3 + 2√2 is positive, and 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172, which is also positive. Therefore, both solutions are valid for Case 1 (x > 0).

Case 2: 2x / (1 + x²) = -1/3.

Multiply both sides by (1 + x²):

2x = - (1 + x²)/3 => 6x = -1 - x² => x² + 6x + 1 = 0.

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

Now, check if these solutions are negative. Let's compute:

-3 + 2√2 ≈ -3 + 2.828 ≈ -0.172, which is negative.

-3 - 2√2 ≈ -3 - 2.828 ≈ -5.828, which is also negative.

Therefore, both solutions are valid for Case 2 (x < 0).

Therefore, combining all solutions, we have four solutions:

From Case 1: 3 + 2√2 and 3 - 2√2.

From Case 2: -3 + 2√2 and -3 - 2√2.

Wait, but hold on. Let me check if these solutions are correct. Let me verify with an example. Let me take x = 3 + 2√2. Let me compute 2x / (1 + x²) and check if it equals 1/3.

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

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.

So 2x / (1 + x²) = (6 + 4√2) / (18 + 12√2). Let me factor numerator and denominator:

Numerator: 2*(3 + 2√2)

Denominator: 6*(3 + 2√2)

Therefore, the ratio is 2/6 = 1/3. Correct. Similarly for x = 3 - 2√2:

x = 3 - 2√2. Compute x²:

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

1 + x² = 18 - 12√2.

2x = 2*(3 - 2√2) = 6 - 4√2.

So 2x / (1 + x²) = (6 - 4√2) / (18 - 12√2) = factor numerator and denominator:

Numerator: 2*(3 - 2√2)

Denominator: 6*(3 - 2√2)

Therefore, ratio is 2/6 = 1/3. Correct.

Similarly for Case 2, take x = -3 + 2√2 ≈ -0.172. Let me compute 2x / (1 + x²):

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

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

1 + x² = 18 - 12√2.

2x = 2*(-3 + 2√2) = -6 + 4√2.

So 2x / (1 + x²) = (-6 + 4√2) / (18 - 12√2). Let me factor numerator and denominator:

Numerator: -2*(3 - 2√2)

Denominator: 6*(3 - 2√2)

Therefore, ratio is -2/6 = -1/3. Which matches Case 2 equation: 2x / (1 + x²) = -1/3. Therefore, correct.

Similarly for x = -3 - 2√2:

x = -3 - 2√2. Compute x²:

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

1 + x² = 18 + 12√2.

2x = 2*(-3 - 2√2) = -6 - 4√2.

So 2x / (1 + x²) = (-6 - 4√2) / (18 + 12√2) = factor numerator and denominator:

Numerator: -2*(3 + 2√2)

Denominator: 6*(3 + 2√2)

Ratio is -2/6 = -1/3. Correct.

Therefore, all four solutions are valid. Therefore, the solutions are:

x = 3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

But let me check if there are any restrictions. For example, when x = 0, but x = 0 is excluded because cot⁻¹(1/x) is undefined. However, our solutions don't include x = 0. Also, for x negative, we have solutions. Therefore, all four solutions are valid.

But let me check if there's any mistake here. Let me check with x = -3 + 2√2. Let me compute tan⁻¹(x) + cot⁻¹(1/x) and check if its sine is 1/3.

Let me compute x = -3 + 2√2 ≈ -3 + 2.828 ≈ -0.172. Let me compute tan⁻¹(x). Let me denote θ = tan⁻¹(x). Then tan θ = x ≈ -0.172. Therefore, θ ≈ -0.17 radians (approx). Then cot⁻¹(1/x) = cot⁻¹(1/(-0.172)) ≈ cot⁻¹(-5.81). Since 1/x ≈ -5.81. As discussed earlier, cot⁻¹(-5.81) = π + tan⁻¹(-1/5.81) ≈ π - tan⁻¹(0.172) ≈ π - 0.17 ≈ 3.0 radians. Therefore, θ + cot⁻¹(1/x) ≈ -0.17 + 3.0 ≈ 2.83 radians. Then sin(2.83) ≈ sin(π - 0.31) ≈ sin(0.31) ≈ 0.304, which is approximately 1/3 ≈ 0.333. Wait, but there's a discrepancy here. Wait, maybe my approximations are too rough. Let me compute more accurately.

Alternatively, since we already derived that sin(A + B) = -2x / (1 + x²) for x < 0, and we have for x = -3 + 2√2, -2x / (1 + x²) = 1/3. Therefore, sin(A + B) = 1/3. Therefore, the exact value is correct. Therefore, even though my approximate calculation gave me something close, it's because of the approximated values. Therefore, the exact solutions are correct.

Therefore, the four solutions are:

x = 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 the same as 2√2 - 3, and -3 - 2√2 is - (3 + 2√2). So maybe writing them in order:

3 + 2√2, 2√2 - 3, - (3 + 2√2), and - (2√2 - 3)? Wait, no:

Wait, -3 + 2√2 is 2√2 - 3, and -3 - 2√2 is -(3 + 2√2). So the four solutions are:

3 + 2√2,

2√2 - 3,

- (3 + 2√2),

and - (2√2 - 3) = -2√2 + 3.

But maybe the problem expects the solutions in a certain order or simplified form. However, the question says "Enter all the solutions, separated by commas." So probably just list them as they are.

But let me check if there are any duplicates or if some solutions are extraneous. For example, when solving quadratics, sometimes extraneous solutions can appear, but in this case, we checked that all four solutions satisfy the original equation. Therefore, they are all valid.

Therefore, the solutions are:

x = 3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

Alternatively, written as:

x = 3 + 2√2, 2√2 - 3, -3 - 2√2, -2√2 - 3? Wait, no:

Wait, -3 - 2√2 is the same as -(3 + 2√2), and -3 + 2√2 is 2√2 - 3. So the four solutions are:

3 + 2√2,

2√2 - 3,

- (3 + 2√2),

and - (2√2 - 3) = -2√2 + 3.

But depending on how you write them, they can be ordered. However, the problem doesn't specify any particular order, so as long as all four are listed, it's okay.

But let me check if there's a mistake here. Wait, when I solved Case 1, I had x² - 6x + 1 = 0, which gave 3 ± 2√2. Then Case 2, x² + 6x + 1 = 0, which gave -3 ± 2√2. Therefore, the four solutions are indeed 3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2. Therefore, these are four distinct solutions.

Therefore, the answer is these four numbers. Let me check if they are all real numbers, which they are. Therefore, the solutions are:

3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

But let me check if there's a possibility that some of these solutions might not satisfy the original equation. For example, let me take x = -3 - 2√2. Let me compute sin(tan⁻¹(x) + cot⁻¹(1/x)).

First, x = -3 - 2√2. Let me compute tan⁻¹(x). Let me denote θ = tan⁻¹(x). Then tan θ = x = -3 - 2√2. Then cot⁻¹(1/x) = cot⁻¹(1/(-3 - 2√2)) = cot⁻¹(-1/(3 + 2√2)). Let me rationalize the denominator:

Multiply numerator and denominator by (3 - 2√2):

-1/(3 + 2√2) * (3 - 2√2)/(3 - 2√2) = - (3 - 2√2)/[(3)^2 - (2√2)^2] = - (3 - 2√2)/(9 - 8) = - (3 - 2√2)/1 = -3 + 2√2.

Therefore, cot⁻¹(-3 + 2√2). Now, since -3 + 2√2 ≈ -3 + 2.828 ≈ -0.172, which is negative. Therefore, cot⁻¹(-0.172) = π + tan⁻¹(-1/0.172) ≈ π - tan⁻¹(5.81). But regardless, using our previous formula, for x < 0, sin(A + B) = -2x / (1 + x²). Let me compute that:

x = -3 - 2√2. Then -2x = 6 + 4√2. 1 + x² = 1 + ( (-3 - 2√2)^2 ) = 1 + 9 + 12√2 + 8 = 18 + 12√2. Therefore, -2x / (1 + x²) = (6 + 4√2)/(18 + 12√2) = same as before, which simplifies to 1/3. Therefore, sin(A + B) = 1/3. Therefore, correct. Therefore, all four solutions are valid.

Therefore, the final answer is these four solutions. Therefore, the solutions are:

3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

But let me check if the problem expects them in a specific order or simplified form. For example, sometimes solutions are written in order from smallest to largest. Let me compute approximate decimal 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

Therefore, ordered from smallest to largest: -5.828, -0.172, 0.172, 5.828. Therefore, the solutions are:

-3 - 2√2, -3 + 2√2, 3 - 2√2, 3 + 2√2.

But the problem says "Enter all the solutions, separated by commas." So as long as all four are included, the order doesn't matter. However, sometimes problems expect the answers in a certain order, but since it's not specified, I can present them in any order.

But let me check if there's a possibility that some solutions are repeated or if there's a mistake in the number of solutions. Wait, the original equation is sin(something) = 1/3. The sine function can take multiple angles to give the same value. However, in our case, we derived that depending on the sign of x, we have different equations. Therefore, leading to four solutions. However, let me check if there's a possibility that some solutions are extraneous. For example, when solving equations involving inverse trigonometric functions, sometimes extraneous solutions can be introduced. However, in our case, we checked that all four solutions satisfy the original equation. Therefore, they are all valid.

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

But let me check once more if there's a different approach that might lead to a different number of solutions. For example, suppose I consider the equation sin(θ) = 1/3, which has solutions θ = arcsin(1/3) + 2πn or π - arcsin(1/3) + 2πn for integer n. However, in our case, θ = tan⁻¹(x) + cot⁻¹(1/x). But we found that θ is either 2 tan⁻¹(x) or 2 tan⁻¹(x) + π. Therefore, depending on the value of x, θ can be in different ranges. However, since tan⁻¹(x) is between -π/2 and π/2, then 2 tan⁻¹(x) is between -π and π. Adding π for the negative x case, θ is between 0 and 2π. However, sin(θ) = 1/3, so θ can be in the first or second quadrant. Therefore, there are two solutions for θ in [0, 2π). However, since θ is determined by x, which can be positive or negative, leading to different equations. Therefore, leading to four solutions for x. Therefore, this seems consistent.

Therefore, I think the four solutions are correct.

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

But since the problem says "Enter all the solutions, separated by commas," and the platform might expect multiple boxed answers or a single box with commas. However, standard practice for such problems is to list all solutions in a single box separated by commas. However, since the user instruction says "put your final answer within \boxed{}", maybe they expect multiple boxed answers? But the original instruction says "Enter all the solutions, separated by commas." So likely, they want all solutions in one box, separated by commas. However, the system might require each solution boxed. But since the user hasn't specified, but in the problem statement, they said "put your final answer within \boxed{}", which usually indicates a single box. However, since there are multiple solutions, maybe they should be in separate boxes. But I need to check the initial problem's instructions again.

Looking back: "Find all solutions to [...] Enter all the solutions, separated by commas. Please reason step by step, and put your final answer within \boxed{}."

So, the user says "put your final answer within \boxed{}". But if there are multiple solutions, how to format? Maybe as multiple boxed expressions separated by commas. However, standard practice in math problems is to list all solutions in a single box, separated by commas. For example, on AoPS or other platforms, multiple solutions are put in one box. Therefore, I will follow that.

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But if the system requires a single box, then:

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But since the user might expect all in one box, maybe:

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But I need to check if the platform allows multiple boxes. Since the original instruction says "put your final answer within \boxed{}", maybe they expect all solutions in one box, separated by commas. For example:

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But actually, in many cases, multiple boxed answers are acceptable. However, since the user hasn't specified, but given the initial problem's structure, I think the correct way is to present all solutions in separate boxes. However, to comply with the instruction "put your final answer within \boxed{}", maybe they want all solutions in one box. For example:

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But I think the standard for such problems on platforms like Brilliant or AoPS is to have multiple boxed answers. However, since the user hasn't specified, but given the initial instruction says "put your final answer within \boxed{}", I might need to check if there's a way to have multiple boxed answers. However, in standard practice here, if there are multiple solutions, they are listed in one box separated by commas. For example:

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But since the user might expect a single box, I'll go with that.

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But actually, looking at the problem again, the original equation is sin(...) = 1/3. The sine function is periodic, but in our case, we derived specific solutions based on the inverse trigonometric functions. Therefore, the four solutions are correct. Therefore, I think the answer should be presented as four boxed expressions.

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \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

Let:
- $ A = \tan^{-1}(x) $
- $ B = \cot^{-1}\left(\frac{1}{x}\right) $

We use the identity:
$$
\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right)
$$
for $ y > 0 $. However, for $ y < 0 $, the identity becomes:
$$
\cot^{-1}(y) = \pi + \tan^{-1}\left(\frac{1}{y}\right)
$$

So, for $ x > 0 $, we have:
$$
\cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(x)
$$
Thus:
$$
\tan^{-1}(x) + \cot^{-1}\left(\frac{1}{x}\right) = 2 \tan^{-1}(x)
$$

For $ x < 0 $, we have:
$$
\cot^{-1}\left(\frac{1}{x}\right) = \pi + \tan^{-1}(x)
$$
So:
$$
\tan^{-1}(x) + \cot^{-1}\left(\frac{1}{x}\right) = 2 \tan^{-1}(x) + \pi
$$

Therefore, the sine of the expression becomes:
- For $ x > 0 $: $ \sin(2 \tan^{-1}(x)) $
- For $ x < 0 $: $ \sin(2 \tan^{-1}(x) + \pi) = -\sin(2 \tan^{-1}(x)) $

---

### Step 2: Use Trigonometric Identity

We use the identity:
$$
\sin(2 \tan^{-1}(x)) = \frac{2x}{1 + x^2}
$$

So the equation becomes:
- For $ x > 0 $: $ \frac{2x}{1 + x^2} = \frac{1}{3} $
- For $ x < 0 $: $ -\frac{2x}{1 + x^2} = \frac{1}{3} \Rightarrow \frac{2x}{1 + x^2} = -\frac{1}{3} $

---

### Step 3: Solve the Quadratic Equations

#### Case 1: $ \frac{2x}{1 + x^2} = \frac{1}{3} $

Multiply both sides by $ 1 + x^2 $:
$$
2x = \frac{1 + x^2}{3} \Rightarrow 6x = 1 + x^2 \Rightarrow x^2 - 6x + 1 = 0
$$

Solve using the quadratic formula:
$$
x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}
$$

Both solutions are positive, so they are valid for $ x > 0 $.

#### Case 2: $ \frac{2x}{1 + x^2} = -\frac{1}{3} $

Multiply both sides by $ 1 + x^2 $:
$$
2x = -\frac{1 + x^2}{3} \Rightarrow 6x = -1 - x^2 \Rightarrow x^2 + 6x + 1 = 0
$$

Solve using the quadratic formula:
$$
x = \frac{-6 \pm \sqrt{36 - 4}}{2} = \frac{-6 \pm \sqrt{32}}{2} = \frac{-6 \pm 4\sqrt{2}}{2} = -3 \pm 2\sqrt{2}
$$

Both solutions are negative, so they are valid for $ x < 0 $.

---

### Step 4: Final Answer

The four solutions to the equation are:

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

[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. Okay, that seems right.

Alternatively, maybe there's a different identity. Let me think. Another way to approach this is to use the identity for sin(A + B). Let me try that approach as well to confirm.

Let me denote A = tan⁻¹(x) and B = cot⁻¹(1/x). Then, the equation is sin(A + B) = 1/3. Let me compute A + B. If A = tan⁻¹(x), then tan A = x. For B = cot⁻¹(1/x), which is the same as tan⁻¹(x) as we discussed earlier. Therefore, B = tan⁻¹(x). Therefore, A + B = 2 tan⁻¹(x). So again, same result. Therefore, sin(2 tan⁻¹(x)) = 1/3.

Alternatively, maybe there's a relationship between A and B? Let me check. If A = tan⁻¹(x), then cot⁻¹(1/x) is equal to π/2 - tan⁻¹(x)? Wait, no. Wait, tan(π/2 - θ) = cot θ. Therefore, cot⁻¹(y) = π/2 - tan⁻¹(y). Wait, so if we have cot⁻¹(1/x), then that would be π/2 - tan⁻¹(1/x). Wait, but earlier I thought cot⁻¹(1/x) = tan⁻¹(x). Let me check with numbers again. Let me take x = 1. Then cot⁻¹(1/1) = cot⁻¹(1) = π/4, and tan⁻¹(1) is also π/4. So that works. If x = 2, then cot⁻¹(1/2) is the angle whose cotangent is 1/2, which is the same as tan⁻¹(2). Because cot(θ) = 1/tan(θ), so if cot(θ) = 1/2, then tan(θ) = 2, so θ = tan⁻¹(2). Therefore, cot⁻¹(1/x) = tan⁻¹(x). Therefore, that seems correct. Therefore, A + B = tan⁻¹(x) + tan⁻¹(x) = 2 tan⁻¹(x). Therefore, sin(2 tan⁻¹(x)) = 1/3.

Alternatively, if I use the identity for sin(A + B), maybe I can compute sin(A + B) directly. Let me try that. Let me compute sin(A + B) where A = tan⁻¹(x) and B = cot⁻¹(1/x). Let me compute sin(A + B) = sin A cos B + cos A sin B. Let me compute sin A, cos A, sin B, cos B.

Since A = tan⁻¹(x), then we can imagine a right triangle where the opposite side is x and adjacent side is 1, so hypotenuse is √(1 + x²). Therefore, sin A = x / √(1 + x²), cos A = 1 / √(1 + x²).

Similarly, B = cot⁻¹(1/x). Let me think about B. cot B = 1/x. So cot B = adjacent / opposite = 1/x. Therefore, if we imagine a right triangle where adjacent side is 1 and opposite side is x, hypotenuse is √(1 + x²). Therefore, sin B = opposite / hypotenuse = x / √(1 + x²), and cos B = adjacent / hypotenuse = 1 / √(1 + x²).

Therefore, sin(A + B) = sin A cos B + cos A sin B = [x / √(1 + x²) * 1 / √(1 + x²)] + [1 / √(1 + x²) * x / √(1 + x²)] = [x / (1 + x²)] + [x / (1 + x²)] = 2x / (1 + x²). Therefore, sin(A + B) = 2x / (1 + x²). Therefore, the original equation is 2x / (1 + x²) = 1/3.

Wait, so that's another way to get the same equation. Therefore, regardless of the approach, we end up with 2x / (1 + x²) = 1/3. Therefore, solving this equation for x.

So let me write that equation:

2x / (1 + x²) = 1/3.

Cross multiplying:

2x * 3 = 1 * (1 + x²)

Which is:

6x = 1 + x²

Rearranging:

x² - 6x + 1 = 0

Wait, hold on. Wait, 6x = 1 + x² => x² - 6x + 1 = 0? Wait, no. Let me check:

Starting from 6x = 1 + x², subtract 6x from both sides:

0 = x² - 6x + 1, so yes, x² - 6x + 1 = 0. Therefore, solving quadratic equation:

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

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

Wait, but hold on. Let me check if there are any restrictions on x. Since tan⁻¹(x) is defined for all real numbers, and cot⁻¹(1/x) is defined when 1/x is defined, so x ≠ 0. Therefore, x cannot be zero. However, our solutions are 3 + 2√2 and 3 - 2√2. Let me check if they are non-zero. 3 - 2√2 is approximately 3 - 2.828 ≈ 0.172, which is positive, so not zero. Therefore, both solutions are valid.

But wait, let me check if there are any other constraints. For example, when dealing with inverse trigonometric functions, sometimes there can be issues with the domain or range. Let me think. The original equation is sin( tan⁻¹(x) + cot⁻¹(1/x) ) = 1/3. We found that tan⁻¹(x) + cot⁻¹(1/x) simplifies to 2 tan⁻¹(x). However, is there a case where tan⁻¹(x) + cot⁻¹(1/x) is not equal to 2 tan⁻¹(x)? Wait, earlier we saw that cot⁻¹(1/x) = tan⁻¹(x). Let me check for negative x. Suppose x is negative. Let me take x = -1. Then tan⁻¹(-1) = -π/4. cot⁻¹(1/(-1)) = cot⁻¹(-1). Now, cot⁻¹(-1) is equal to π - cot⁻¹(1) = π - π/4 = 3π/4? Wait, but cot⁻¹(y) is usually defined to have a range of (0, π). So cot⁻¹(-1) would be π - cot⁻¹(1) = π - π/4 = 3π/4. However, tan⁻¹(x) for x negative is negative. So in this case, tan⁻¹(-1) + cot⁻¹(-1) = -π/4 + 3π/4 = π/2. However, 2 tan⁻¹(-1) = 2*(-π/4) = -π/2. But sin(-π/2) = -1, whereas sin(π/2) = 1. Therefore, there's a discrepancy here. Therefore, my previous conclusion that cot⁻¹(1/x) = tan⁻¹(x) is only valid for positive x? Wait, so maybe there's a mistake here.

Wait, so if x is negative, then cot⁻¹(1/x) is not equal to tan⁻¹(x). Let me check again. Let me recall that cot⁻¹(y) is equal to tan⁻¹(1/y) when y > 0, but for y < 0, cot⁻¹(y) is equal to tan⁻¹(1/y) + π? Or something like that? Wait, let me recall the exact ranges.

The range of cot⁻¹(y) is typically (0, π). The range of tan⁻¹(y) is (-π/2, π/2). So if we have cot⁻¹(y) = π/2 - tan⁻¹(y) for y > 0. Wait, let me check. For y > 0, cot⁻¹(y) = tan⁻¹(1/y). But tan⁻¹(1/y) is in (-π/2, π/2). However, cot⁻¹(y) is in (0, π). Therefore, for y > 0, cot⁻¹(y) = π/2 - tan⁻¹(y). Wait, let me check with y = 1. cot⁻¹(1) = π/4, and π/2 - tan⁻¹(1) = π/2 - π/4 = π/4. Correct. For y = 2, cot⁻¹(2) = π/2 - tan⁻¹(2). So that's correct. However, for y < 0, cot⁻¹(y) is in (0, π). Let me take y = -1. Then cot⁻¹(-1) is π - cot⁻¹(1) = π - π/4 = 3π/4. On the other hand, tan⁻¹(1/y) = tan⁻¹(-1) = -π/4. Therefore, cot⁻¹(y) = π + tan⁻¹(1/y) for y < 0? Wait, π + tan⁻¹(1/y) would be π + (-π/4) = 3π/4, which matches. Therefore, in general, cot⁻¹(y) = tan⁻¹(1/y) if y > 0, and cot⁻¹(y) = π + tan⁻¹(1/y) if y < 0. Therefore, in general, cot⁻¹(y) = tan⁻¹(1/y) + π * H(y), where H(y) is 0 if y > 0 and 1 if y < 0? Wait, not sure. But in any case, for our problem, we have cot⁻¹(1/x). Let me consider two cases: x > 0 and x < 0.

Case 1: x > 0. Then cot⁻¹(1/x) = tan⁻¹(x). Therefore, tan⁻¹(x) + cot⁻¹(1/x) = tan⁻¹(x) + tan⁻¹(x) = 2 tan⁻¹(x). Therefore, sin(2 tan⁻¹(x)) = 1/3.

Case 2: x < 0. Then cot⁻¹(1/x) = π + tan⁻¹(x). Wait, let me check. If x < 0, then 1/x < 0. Therefore, cot⁻¹(1/x) = π + tan⁻¹(1/(1/x))? Wait, no. Wait, earlier we saw that for y < 0, cot⁻¹(y) = π + tan⁻¹(1/y). Wait, no. Wait, for y < 0, cot⁻¹(y) = π + tan⁻¹(1/y)? Let me check with y = -1. cot⁻¹(-1) = 3π/4. tan⁻¹(1/y) = tan⁻¹(-1) = -π/4. So π + tan⁻¹(1/y) = π - π/4 = 3π/4. Correct. Therefore, for y < 0, cot⁻¹(y) = π + tan⁻¹(1/y). Therefore, in our case, cot⁻¹(1/x) where x < 0, so 1/x < 0. Therefore, cot⁻¹(1/x) = π + tan⁻¹(1/(1/x)) = π + tan⁻¹(x). Therefore, tan⁻¹(x) + cot⁻¹(1/x) = tan⁻¹(x) + π + tan⁻¹(x) = 2 tan⁻¹(x) + π. Therefore, sin(2 tan⁻¹(x) + π) = sin(2 tan⁻¹(x)) * cos(π) + cos(2 tan⁻¹(x)) * sin(π) = -sin(2 tan⁻¹(x)) + 0 = -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, depending on the sign of x, we have different equations. Therefore, my previous approach only considered x > 0, but we need to check both cases.

Therefore, let me re-examine the problem. So, the original equation is:

sin( tan⁻¹(x) + cot⁻¹(1/x) ) = 1/3.

But depending on the sign of x, the expression inside the sine function is either 2 tan⁻¹(x) or 2 tan⁻¹(x) + π. Therefore, we need to consider both cases.

Case 1: x > 0. Then, sin(2 tan⁻¹(x)) = 1/3.

Case 2: x < 0. Then, sin(2 tan⁻¹(x) + π) = 1/3 => -sin(2 tan⁻¹(x)) = 1/3 => sin(2 tan⁻¹(x)) = -1/3.

Therefore, we need to solve both equations:

For x > 0: sin(2 tan⁻¹(x)) = 1/3.

For x < 0: sin(2 tan⁻¹(x)) = -1/3.

But earlier, when I considered the general case, I derived that sin(A + B) = 2x / (1 + x²). Wait, let me check that again. Earlier, I used the identity for sin(A + B) and found that sin(A + B) = 2x / (1 + x²). However, that was under the assumption that A = tan⁻¹(x) and B = cot⁻¹(1/x). But if x is negative, does that formula still hold?

Wait, let me recalculate sin(A + B) for x negative. Let me take x negative. Let me take x = -k where k > 0. Then tan⁻¹(x) = tan⁻¹(-k) = -tan⁻¹(k). Similarly, cot⁻¹(1/x) = cot⁻¹(-1/k). As we saw earlier, cot⁻¹(-1/k) = π + tan⁻¹(-k) = π - tan⁻¹(k). Therefore, A + B = tan⁻¹(x) + cot⁻¹(1/x) = -tan⁻¹(k) + π - tan⁻¹(k) = π - 2 tan⁻¹(k). Therefore, sin(A + B) = sin(π - 2 tan⁻¹(k)) = sin(2 tan⁻¹(k)) because sin(π - θ) = sin θ. Therefore, sin(A + B) = sin(2 tan⁻¹(k)) = 2k / (1 + k²). But since x = -k, then k = -x. Therefore, sin(A + B) = 2*(-x) / (1 + x²). Wait, because k = -x, so 2k / (1 + k²) = 2*(-x) / (1 + x²). Therefore, sin(A + B) = -2x / (1 + x²). Therefore, in this case, sin(A + B) = -2x / (1 + x²). Therefore, the original equation sin(A + B) = 1/3 becomes -2x / (1 + x²) = 1/3. Therefore, for x < 0, we have -2x / (1 + x²) = 1/3. Therefore, combining both cases:

For x > 0: 2x / (1 + x²) = 1/3.

For x < 0: -2x / (1 + x²) = 1/3.

Therefore, solving both equations.

So let me write both equations:

Case 1: 2x / (1 + x²) = 1/3.

Case 2: -2x / (1 + x²) = 1/3 => 2x / (1 + x²) = -1/3.

Therefore, solving both equations.

Starting with Case 1: 2x / (1 + x²) = 1/3.

Multiply both sides by (1 + x²):

2x = (1 + x²)/3 => 6x = 1 + x² => x² - 6x + 1 = 0.

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

Now, check if these solutions are positive. 3 + 2√2 is positive, and 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172, which is also positive. Therefore, both solutions are valid for Case 1 (x > 0).

Case 2: 2x / (1 + x²) = -1/3.

Multiply both sides by (1 + x²):

2x = - (1 + x²)/3 => 6x = -1 - x² => x² + 6x + 1 = 0.

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

Now, check if these solutions are negative. Let's compute:

-3 + 2√2 ≈ -3 + 2.828 ≈ -0.172, which is negative.

-3 - 2√2 ≈ -3 - 2.828 ≈ -5.828, which is also negative.

Therefore, both solutions are valid for Case 2 (x < 0).

Therefore, combining all solutions, we have four solutions:

From Case 1: 3 + 2√2 and 3 - 2√2.

From Case 2: -3 + 2√2 and -3 - 2√2.

Wait, but hold on. Let me check if these solutions are correct. Let me verify with an example. Let me take x = 3 + 2√2. Let me compute 2x / (1 + x²) and check if it equals 1/3.

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

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.

So 2x / (1 + x²) = (6 + 4√2) / (18 + 12√2). Let me factor numerator and denominator:

Numerator: 2*(3 + 2√2)

Denominator: 6*(3 + 2√2)

Therefore, the ratio is 2/6 = 1/3. Correct. Similarly for x = 3 - 2√2:

x = 3 - 2√2. Compute x²:

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

1 + x² = 18 - 12√2.

2x = 2*(3 - 2√2) = 6 - 4√2.

So 2x / (1 + x²) = (6 - 4√2) / (18 - 12√2) = factor numerator and denominator:

Numerator: 2*(3 - 2√2)

Denominator: 6*(3 - 2√2)

Therefore, ratio is 2/6 = 1/3. Correct.

Similarly for Case 2, take x = -3 + 2√2 ≈ -0.172. Let me compute 2x / (1 + x²):

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

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

1 + x² = 18 - 12√2.

2x = 2*(-3 + 2√2) = -6 + 4√2.

So 2x / (1 + x²) = (-6 + 4√2) / (18 - 12√2). Let me factor numerator and denominator:

Numerator: -2*(3 - 2√2)

Denominator: 6*(3 - 2√2)

Therefore, ratio is -2/6 = -1/3. Which matches Case 2 equation: 2x / (1 + x²) = -1/3. Therefore, correct.

Similarly for x = -3 - 2√2:

x = -3 - 2√2. Compute x²:

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

1 + x² = 18 + 12√2.

2x = 2*(-3 - 2√2) = -6 - 4√2.

So 2x / (1 + x²) = (-6 - 4√2) / (18 + 12√2) = factor numerator and denominator:

Numerator: -2*(3 + 2√2)

Denominator: 6*(3 + 2√2)

Ratio is -2/6 = -1/3. Correct.

Therefore, all four solutions are valid. Therefore, the solutions are:

x = 3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

But let me check if there are any restrictions. For example, when x = 0, but x = 0 is excluded because cot⁻¹(1/x) is undefined. However, our solutions don't include x = 0. Also, for x negative, we have solutions. Therefore, all four solutions are valid.

But let me check if there's any mistake here. Let me check with x = -3 + 2√2. Let me compute tan⁻¹(x) + cot⁻¹(1/x) and check if its sine is 1/3.

Let me compute x = -3 + 2√2 ≈ -3 + 2.828 ≈ -0.172. Let me compute tan⁻¹(x). Let me denote θ = tan⁻¹(x). Then tan θ = x ≈ -0.172. Therefore, θ ≈ -0.17 radians (approx). Then cot⁻¹(1/x) = cot⁻¹(1/(-0.172)) ≈ cot⁻¹(-5.81). Since 1/x ≈ -5.81. As discussed earlier, cot⁻¹(-5.81) = π + tan⁻¹(-1/5.81) ≈ π - tan⁻¹(0.172) ≈ π - 0.17 ≈ 3.0 radians. Therefore, θ + cot⁻¹(1/x) ≈ -0.17 + 3.0 ≈ 2.83 radians. Then sin(2.83) ≈ sin(π - 0.31) ≈ sin(0.31) ≈ 0.304, which is approximately 1/3 ≈ 0.333. Wait, but there's a discrepancy here. Wait, maybe my approximations are too rough. Let me compute more accurately.

Alternatively, since we already derived that sin(A + B) = -2x / (1 + x²) for x < 0, and we have for x = -3 + 2√2, -2x / (1 + x²) = 1/3. Therefore, sin(A + B) = 1/3. Therefore, the exact value is correct. Therefore, even though my approximate calculation gave me something close, it's because of the approximated values. Therefore, the exact solutions are correct.

Therefore, the four solutions are:

x = 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 the same as 2√2 - 3, and -3 - 2√2 is - (3 + 2√2). So maybe writing them in order:

3 + 2√2, 2√2 - 3, - (3 + 2√2), and - (2√2 - 3)? Wait, no:

Wait, -3 + 2√2 is 2√2 - 3, and -3 - 2√2 is -(3 + 2√2). So the four solutions are:

3 + 2√2,

2√2 - 3,

- (3 + 2√2),

and - (2√2 - 3) = -2√2 + 3.

But maybe the problem expects the solutions in a certain order or simplified form. However, the question says "Enter all the solutions, separated by commas." So probably just list them as they are.

But let me check if there are any duplicates or if some solutions are extraneous. For example, when solving quadratics, sometimes extraneous solutions can appear, but in this case, we checked that all four solutions satisfy the original equation. Therefore, they are all valid.

Therefore, the solutions are:

x = 3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

Alternatively, written as:

x = 3 + 2√2, 2√2 - 3, -3 - 2√2, -2√2 - 3? Wait, no:

Wait, -3 - 2√2 is the same as -(3 + 2√2), and -3 + 2√2 is 2√2 - 3. So the four solutions are:

3 + 2√2,

2√2 - 3,

- (3 + 2√2),

and - (2√2 - 3) = -2√2 + 3.

But depending on how you write them, they can be ordered. However, the problem doesn't specify any particular order, so as long as all four are listed, it's okay.

But let me check if there's a mistake here. Wait, when I solved Case 1, I had x² - 6x + 1 = 0, which gave 3 ± 2√2. Then Case 2, x² + 6x + 1 = 0, which gave -3 ± 2√2. Therefore, the four solutions are indeed 3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2. Therefore, these are four distinct solutions.

Therefore, the answer is these four numbers. Let me check if they are all real numbers, which they are. Therefore, the solutions are:

3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

But let me check if there's a possibility that some of these solutions might not satisfy the original equation. For example, let me take x = -3 - 2√2. Let me compute sin(tan⁻¹(x) + cot⁻¹(1/x)).

First, x = -3 - 2√2. Let me compute tan⁻¹(x). Let me denote θ = tan⁻¹(x). Then tan θ = x = -3 - 2√2. Then cot⁻¹(1/x) = cot⁻¹(1/(-3 - 2√2)) = cot⁻¹(-1/(3 + 2√2)). Let me rationalize the denominator:

Multiply numerator and denominator by (3 - 2√2):

-1/(3 + 2√2) * (3 - 2√2)/(3 - 2√2) = - (3 - 2√2)/[(3)^2 - (2√2)^2] = - (3 - 2√2)/(9 - 8) = - (3 - 2√2)/1 = -3 + 2√2.

Therefore, cot⁻¹(-3 + 2√2). Now, since -3 + 2√2 ≈ -3 + 2.828 ≈ -0.172, which is negative. Therefore, cot⁻¹(-0.172) = π + tan⁻¹(-1/0.172) ≈ π - tan⁻¹(5.81). But regardless, using our previous formula, for x < 0, sin(A + B) = -2x / (1 + x²). Let me compute that:

x = -3 - 2√2. Then -2x = 6 + 4√2. 1 + x² = 1 + ( (-3 - 2√2)^2 ) = 1 + 9 + 12√2 + 8 = 18 + 12√2. Therefore, -2x / (1 + x²) = (6 + 4√2)/(18 + 12√2) = same as before, which simplifies to 1/3. Therefore, sin(A + B) = 1/3. Therefore, correct. Therefore, all four solutions are valid.

Therefore, the final answer is these four solutions. Therefore, the solutions are:

3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

But let me check if the problem expects them in a specific order or simplified form. For example, sometimes solutions are written in order from smallest to largest. Let me compute approximate decimal 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

Therefore, ordered from smallest to largest: -5.828, -0.172, 0.172, 5.828. Therefore, the solutions are:

-3 - 2√2, -3 + 2√2, 3 - 2√2, 3 + 2√2.

But the problem says "Enter all the solutions, separated by commas." So as long as all four are included, the order doesn't matter. However, sometimes problems expect the answers in a certain order, but since it's not specified, I can present them in any order.

But let me check if there's a possibility that some solutions are repeated or if there's a mistake in the number of solutions. Wait, the original equation is sin(something) = 1/3. The sine function can take multiple angles to give the same value. However, in our case, we derived that depending on the sign of x, we have different equations. Therefore, leading to four solutions. However, let me check if there's a possibility that some solutions are extraneous. For example, when solving equations involving inverse trigonometric functions, sometimes extraneous solutions can be introduced. However, in our case, we checked that all four solutions satisfy the original equation. Therefore, they are all valid.

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

But let me check once more if there's a different approach that might lead to a different number of solutions. For example, suppose I consider the equation sin(θ) = 1/3, which has solutions θ = arcsin(1/3) + 2πn or π - arcsin(1/3) + 2πn for integer n. However, in our case, θ = tan⁻¹(x) + cot⁻¹(1/x). But we found that θ is either 2 tan⁻¹(x) or 2 tan⁻¹(x) + π. Therefore, depending on the value of x, θ can be in different ranges. However, since tan⁻¹(x) is between -π/2 and π/2, then 2 tan⁻¹(x) is between -π and π. Adding π for the negative x case, θ is between 0 and 2π. However, sin(θ) = 1/3, so θ can be in the first or second quadrant. Therefore, there are two solutions for θ in [0, 2π). However, since θ is determined by x, which can be positive or negative, leading to different equations. Therefore, leading to four solutions for x. Therefore, this seems consistent.

Therefore, I think the four solutions are correct.

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

But since the problem says "Enter all the solutions, separated by commas," and the platform might expect multiple boxed answers or a single box with commas. However, standard practice for such problems is to list all solutions in a single box separated by commas. However, since the user instruction says "put your final answer within \boxed{}", maybe they expect multiple boxed answers? But the original instruction says "Enter all the solutions, separated by commas." So likely, they want all solutions in one box, separated by commas. However, the system might require each solution boxed. But since the user hasn't specified, but in the problem statement, they said "put your final answer within \boxed{}", which usually indicates a single box. However, since there are multiple solutions, maybe they should be in separate boxes. But I need to check the initial problem's instructions again.

Looking back: "Find all solutions to [...] Enter all the solutions, separated by commas. Please reason step by step, and put your final answer within \boxed{}."

So, the user says "put your final answer within \boxed{}". But if there are multiple solutions, how to format? Maybe as multiple boxed expressions separated by commas. However, standard practice in math problems is to list all solutions in a single box, separated by commas. For example, on AoPS or other platforms, multiple solutions are put in one box. Therefore, I will follow that.

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But if the system requires a single box, then:

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But since the user might expect all in one box, maybe:

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But I need to check if the platform allows multiple boxes. Since the original instruction says "put your final answer within \boxed{}", maybe they expect all solutions in one box, separated by commas. For example:

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But actually, in many cases, multiple boxed answers are acceptable. However, since the user hasn't specified, but given the initial problem's structure, I think the correct way is to present all solutions in separate boxes. However, to comply with the instruction "put your final answer within \boxed{}", maybe they want all solutions in one box. For example:

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But I think the standard for such problems on platforms like Brilliant or AoPS is to have multiple boxed answers. However, since the user hasn't specified, but given the initial instruction says "put your final answer within \boxed{}", I might need to check if there's a way to have multiple boxed answers. However, in standard practice here, if there are multiple solutions, they are listed in one box separated by commas. For example:

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But since the user might expect a single box, I'll go with that.

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{-3 - 2\sqrt{2}}

But actually, looking at the problem again, the original equation is sin(...) = 1/3. The sine function is periodic, but in our case, we derived specific solutions based on the inverse trigonometric functions. Therefore, the four solutions are correct. Therefore, I think the answer should be presented as four boxed expressions.

**Final Answer**
\boxed{3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \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

Let:
- $ A = \tan^{-1}(x) $
- $ B = \cot^{-1}\left(\frac{1}{x}\right) $

We use the identity:
$$
\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right)
$$
for $ y > 0 $. However, for $ y < 0 $, the identity becomes:
$$
\cot^{-1}(y) = \pi + \tan^{-1}\left(\frac{1}{y}\right)
$$

So, for $ x > 0 $, we have:
$$
\cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(x)
$$
Thus:
$$
\tan^{-1}(x) + \cot^{-1}\left(\frac{1}{x}\right) = 2 \tan^{-1}(x)
$$

For $ x < 0 $, we have:
$$
\cot^{-1}\left(\frac{1}{x}\right) = \pi + \tan^{-1}(x)
$$
So:
$$
\tan^{-1}(x) + \cot^{-1}\left(\frac{1}{x}\right) = 2 \tan^{-1}(x) + \pi
$$

Therefore, the sine of the expression becomes:
- For $ x > 0 $: $ \sin(2 \tan^{-1}(x)) $
- For $ x < 0 $: $ \sin(2 \tan^{-1}(x) + \pi) = -\sin(2 \tan^{-1}(x)) $

---

### Step 2: Use Trigonometric Identity

We use the identity:
$$
\sin(2 \tan^{-1}(x)) = \frac{2x}{1 + x^2}
$$

So the equation becomes:
- For $ x > 0 $: $ \frac{2x}{1 + x^2} = \frac{1}{3} $
- For $ x < 0 $: $ -\frac{2x}{1 + x^2} = \frac{1}{3} \Rightarrow \frac{2x}{1 + x^2} = -\frac{1}{3} $

---

### Step 3: Solve the Quadratic Equations

#### Case 1: $ \frac{2x}{1 + x^2} = \frac{1}{3} $

Multiply both sides by $ 1 + x^2 $:
$$
2x = \frac{1 + x^2}{3} \Rightarrow 6x = 1 + x^2 \Rightarrow x^2 - 6x + 1 = 0
$$

Solve using the quadratic formula:
$$
x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}
$$

Both solutions are positive, so they are valid for $ x > 0 $.

#### Case 2: $ \frac{2x}{1 + x^2} = -\frac{1}{3} $

Multiply both sides by $ 1 + x^2 $:
$$
2x = -\frac{1 + x^2}{3} \Rightarrow 6x = -1 - x^2 \Rightarrow x^2 + 6x + 1 = 0
$$

Solve using the quadratic formula:
$$
x = \frac{-6 \pm \sqrt{36 - 4}}{2} = \frac{-6 \pm \sqrt{32}}{2} = \frac{-6 \pm 4\sqrt{2}}{2} = -3 \pm 2\sqrt{2}
$$

Both solutions are negative, so they are valid for $ x < 0 $.

---

### Step 4: Final Answer

The four solutions to the equation are:

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

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