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

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

First, I remember that inverse trigonometric functions can sometimes be tricky, but maybe there's a relationship between $\tan^{-1}(x)$ and $\cot^{-1}(1/x)$. Let me think... I recall that $\cot^{-1}(y)$ is equal to $\tan^{-1}(1/y)$ when $y > 0$. Is that right? Let me verify. If I consider $\theta = \cot^{-1}(y)$, then $\cot(\theta) = y$, which means $\tan(\theta) = 1/y$, so $\theta = \tan^{-1}(1/y)$. Yeah, that seems correct. So, if $y = 1/x$, then $\cot^{-1}(1/x) = \tan^{-1}(x)$. Wait, but only if $1/x > 0$, which would mean $x > 0$. Hmm, but what if $x$ is negative? Then $1/x$ would be negative, and $\cot^{-1}(1/x)$ would be in a different quadrant? Let me think about that.

But before getting bogged down with the domain issues, maybe I can just assume $x > 0$ first and see where that leads me. Maybe there are solutions for negative $x$ as well? Let's check.

So, if $x > 0$, then $\cot^{-1}(1/x) = \tan^{-1}(x)$. Therefore, the argument inside the sine function becomes $\tan^{-1}(x) + \tan^{-1}(x) = 2\tan^{-1}(x)$. Therefore, the equation simplifies to:

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

Okay, that seems manageable. Let me work with that. Let me denote $\theta = \tan^{-1}(x)$, so that $\theta$ is in the first quadrant because $x > 0$. Then the equation becomes:

$$
\sin(2\theta) = \frac{1}{3}.
$$

Using the double-angle identity, $\sin(2\theta) = 2\sin\theta \cos\theta$. Also, since $\theta = \tan^{-1}(x)$, we can think of a right triangle where the opposite side is $x$ and adjacent is 1, so hypotenuse is $\sqrt{1 + x^2}$. Therefore, $\sin\theta = \frac{x}{\sqrt{1 + x^2}}$ and $\cos\theta = \frac{1}{\sqrt{1 + x^2}}$. Therefore:

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

Therefore, the equation becomes:

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

Cross-multiplying:

$$
6x = 1 + x^2
$$
$$
x^2 - 6x + 1 = 0.
$$

Solving this quadratic equation:

$$
x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 1}}{2} = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}.
$$

So, $x = 3 + 2\sqrt{2}$ or $x = 3 - 2\sqrt{2}$. Let me check if these are positive. Since $\sqrt{2} \approx 1.414, so 2√2 ≈ 2.828. Therefore, 3 - 2.828 ≈ 0.172, which is positive. So both solutions are positive, so they are valid under our initial assumption that $x > 0$.

But wait, what if $x$ is negative? Let me check if there could be solutions for negative $x$. Let me consider $x < 0$. Then, $\cot^{-1}(1/x)$... Let me recall that $\cot^{-1}(y)$ is defined for all real numbers, and it returns values in (0, π). If $y$ is negative, $\cot^{-1}(y)$ would be in (π/2, π). However, $\tan^{-1}(x)$ for $x < 0$ would be in (-π/2, 0).

But if I try to relate $\tan^{-1}(x)$ and $\cot^{-1}(1/x)$, when $x < 0$, is there a relationship? Let me think. Let me take $x = -a$ where $a > 0$. Then $\cot^{-1}(1/x) = \cot^{-1}(-1/a)$. But $\cot^{-1}(-1/a)$ is equal to π - $\cot^{-1}(1/a)$. Because cotangent is negative in the second quadrant. So, $\cot^{-1}(-y) = π - \cot^{-1}(y)$ for $y > 0$. Therefore, $\cot^{-1}(-1/a) = π - \cot^{-1}(1/a)$.

On the other hand, $\tan^{-1}(x) = \tan^{-1}(-a) = -\tan^{-1}(a)$.

Therefore, the argument inside the sine function is:

$$
\tan^{-1}(x) + \cot^{-1}(1/x) = -\tan^{-1}(a) + π - \cot^{-1}(1/a).
$$

But earlier, for positive $a$, we had $\cot^{-1}(1/a) = \tan^{-1}(a)$. So substituting that in:

$$
- \tan^{-1}(a) + π - \tan^{-1}(a) = π - 2 \tan^{-1}(a).
$$

So, the equation becomes:

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

But $\sin(π - θ) = \sin θ$, so this simplifies to:

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

Wait, that's the same equation as before! So even when $x$ is negative, we end up with the same equation as when $x$ is positive. Therefore, solving $\sin(2θ) = 1/3$ would give us solutions for both positive and negative $x$. However, we need to check if there are additional constraints.

But let me think again. If $x$ is negative, then $a = -x > 0$, so solving for $a$ gives the same quadratic equation as before. So:

$$
\frac{2a}{1 + a^2} = \frac{1}{3} \implies a^2 - 6a + 1 = 0 \implies a = 3 \pm 2\sqrt{2}.
$$

Since $a > 0$, both solutions are valid. Therefore, $x = -a$ would be $x = - (3 + 2\sqrt{2})$ and $x = - (3 - 2\sqrt{2})$.

Wait a second, but we need to check if these solutions actually satisfy the original equation. Because sometimes when dealing with inverse trigonometric functions, especially with different quadrants, we might get extraneous solutions.

But let's think again. For the case where $x$ is negative, we had:

$$
\sin\left( π - 2 \tan^{-1}(a) \right) = \sin(2 \tan^{-1}(a)) = 1/3.
$$

But since $\sin(π - θ) = \sin θ$, and we already have that equation. Therefore, the equation for $a$ is the same as for positive $x$, so the solutions for $a$ are $3 \pm 2\sqrt{2}$, so $x = - (3 \pm 2\sqrt{2})$.

But wait, but we have to check if these values are valid. Let me check with $x = 3 + 2\sqrt{2}$, which is positive. Let me compute the original expression. But maybe there's something else. Let me check if the argument inside the sine function is within the correct range.

For positive $x$, the argument is $2 \tan^{-1}(x)$. Since $x > 0$, $\tan^{-1}(x)$ is between 0 and π/2, so doubling it would be between 0 and π. Therefore, the sine of that is between 0 and 1, which is okay.

For negative $x$, the argument was $π - 2 \tan^{-1}(a)$, where $a > 0$. Since $a > 0$, $\tan^{-1}(a)$ is between 0 and π/2, so $2 \tan^{-1}(a)$ is between 0 and π, so π - 2 tan^{-1}(a) is between 0 and π. Therefore, the sine is also between 0 and 1, which is okay. Therefore, both positive and negative solutions are valid.

Therefore, the solutions are $x = 3 + 2\sqrt{2}$, $x = 3 - 2\sqrt{2}$, $x = - (3 + 2\sqrt{2})$, and $x = - (3 - 2\sqrt{2})$?

Wait, but hold on. Let me check if there are only two solutions or four solutions. Wait, the equation $\sin(\theta) = 1/3$ has two solutions in [0, π], which is the range we have for the argument of the sine function (either $2 \tan^{-1}(x)$ or $π - 2 \tan^{-1}(a)$). So, if we consider the general solution for $\sin(\theta) = 1/3$, we have $\theta = \arcsin(1/3)$ or $\theta = π - \arcsin(1/3)$.

But in our case, for positive $x$, the argument is $2 \tan^{-1}(x)$, which is in (0, π). Therefore, solving $2 \tan^{-1}(x) = \arcsin(1/3)$ or $2 \tan^{-1}(x) = π - \arcsin(1/3)$. Wait, but earlier when I set up the equation, I had $\sin(2 \tan^{-1}(x)) = 1/3$, which would correspond to $2 \tan^{-1}(x)$ being in (0, π). Therefore, it's possible that there are two solutions for $x$ in positive numbers. Similarly for negative $x$, but since we ended up with same equation, but $x$ negative. However, in our earlier step, when solving the quadratic, we found two solutions for $a$, which led to two solutions for $x$. So maybe there are only two solutions? Wait, maybe I confused something here.

Wait, let me backtrack. When I considered $x > 0$, I had the equation $\frac{2x}{1 + x^2} = 1/3$, which led to $x^2 - 6x + 1 = 0$, whose solutions are $x = 3 \pm 2\sqrt{2}$. Both positive, so both valid for $x > 0$. Then for $x < 0$, we found that the equation reduces to the same $\frac{2a}{1 + a^2} = 1/3$, with $a = -x > 0$, which gives $a = 3 \pm 2\sqrt{2}$, so $x = -a$ is $x = -3 \pm 2\sqrt{2}$. However, $-3 - 2\sqrt{2}$ is negative, and $-3 + 2\sqrt{2}$ is approximately -3 + 2.828 ≈ -0.172, which is also negative.

Therefore, in total, there are four solutions: $3 + 2\sqrt{2}$, $3 - 2\sqrt{2}$, $-3 - 2\sqrt{2}$, and $-3 + 2\sqrt{2}$. Wait, but hold on, maybe not? Let me check.

Wait, if we have four solutions, but perhaps some of them are extraneous? Let me check with specific values.

Let me first check $x = 3 + 2\sqrt{2}$. Let's compute $\tan^{-1}(x) + \cot^{-1}(1/x)$. Since $x > 0$, $\cot^{-1}(1/x) = \tan^{-1}(x)$. Therefore, the sum is $2 \tan^{-1}(x)$. Then $\sin(2 \tan^{-1}(x))$. As we had earlier, this equals $2x/(1 + x^2)$. Let's compute that with $x = 3 + 2\sqrt{2}$:

First compute $x = 3 + 2\sqrt{2}$. Then $1 + x^2 = 1 + (3 + 2\sqrt{2})^2$. Let me compute that:

$ (3 + 2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} $. So $1 + x^2 = 18 + 12\sqrt{2}$. Then $2x = 2*(3 + 2\sqrt{2}) = 6 + 4\sqrt{2}$. Therefore:

$ \frac{2x}{1 + x^2} = \frac{6 + 4\sqrt{2}}{18 + 12\sqrt{2}} $. Let's rationalize the denominator. Multiply numerator and denominator by the conjugate of the denominator's denominator? Wait, denominator is 18 + 12√2. Alternatively, factor numerator and denominator:

Numerator: 6 + 4√2 = 2*(3 + 2√2)

Denominator: 18 + 12√2 = 6*(3 + 2√2)

Therefore, the ratio is 2*(3 + 2√2)/[6*(3 + 2√2)] = 2/6 = 1/3. Which matches. So yes, that's correct.

Similarly for $x = 3 - 2\sqrt{2}$. Let's check that:

x = 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172. Then 1 + x^2:

x² = (3 - 2√2)^2 = 9 - 12√2 + 8 = 17 - 12√2. Therefore, 1 + x² = 18 - 12√2. Then 2x = 6 - 4√2. Therefore:

(6 - 4√2)/(18 - 12√2). Again, factor numerator and denominator:

Numerator: 2*(3 - 2√2)

Denominator: 6*(3 - 2√2)

So ratio is 2/6 = 1/3. Correct.

Now check the negative solutions. Let's take x = -3 - 2√2. Let me compute $\tan^{-1}(x) + \cot^{-1}(1/x)$. Since x is negative, let me use the previous analysis. Let me denote a = -x = 3 + 2√2. Then:

$\tan^{-1}(x) + \cot^{-1}(1/x) = -\tan^{-1}(a) + π - \cot^{-1}(1/a)$. But since a > 0, $\cot^{-1}(1/a) = \tan^{-1}(a)$. Therefore:

- tan^{-1}(a) + π - tan^{-1}(a) = π - 2 tan^{-1}(a). Then sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3, as before. So same result. Therefore, the value is 1/3. Therefore, x = -a = -3 - 2√2 is valid.

Similarly for x = -3 + 2√2. Let me check that. Let a = -x = 3 - 2√2. Then the same steps apply. So sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3. Which again, for a = 3 - 2√2, we already saw that 2a/(1 + a²) = 1/3. Therefore, this also holds. Therefore, all four solutions are valid.

But let me check if all four solutions are distinct. Let me compute approximate numerical values:

3 + 2√2 ≈ 3 + 2.828 ≈ 5.828

3 - 2√2 ≈ 0.172

-3 - 2√2 ≈ -5.828

-3 + 2√2 ≈ -0.172

Yes, all four are distinct. Therefore, there are four solutions. However, I need to check if the original equation allows all four. Wait, but maybe there's a mistake here. Let me check for x negative. Let me pick x = -0.172, which is - (3 - 2√2). Let me compute the original expression:

sin( tan^{-1}(x) + cot^{-1}(1/x) )

First compute tan^{-1}(x). Since x is negative, tan^{-1}(x) is in (-π/2, 0). Let me compute tan^{-1}(-0.172). Let me denote that as θ. Then cot^{-1}(1/x) = cot^{-1}(1/(-0.172)) = cot^{-1}(-5.828). But cot^{-1}(-5.828) is equal to π - cot^{-1}(5.828). Let me compute cot^{-1}(5.828). Since cot^{-1}(y) is equal to tan^{-1}(1/y), so cot^{-1}(5.828) is tan^{-1}(1/5.828) ≈ tan^{-1}(0.172) ≈ θ' where tan θ' ≈ 0.172, which is approximately θ, since tan(θ) = -0.172. But θ is negative, so θ' is positive. Therefore, cot^{-1}(-5.828) = π - θ'.

So tan^{-1}(x) + cot^{-1}(1/x) = θ + π - θ' = π + θ - θ'. But θ = tan^{-1}(x) = tan^{-1}(-0.172) ≈ -θ', since tan(-θ') = -tan(theta') ≈ -0.172. Therefore, θ ≈ -θ', so θ - θ' ≈ -θ' - θ' = -2θ'. Therefore, the total expression is π - 2θ', which is π - 2 tan^{-1}(a), with a = 3 - 2√2 ≈ 0.172. Therefore, sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3. So that works.

Similarly, for the other negative value. Therefore, all four solutions are valid. Therefore, the answer should have four solutions: 3 + 2√2, 3 - 2√2, -3 - 2√2, -3 + 2√2.

But let me check if there's any other constraints. For example, is there any domain restrictions on the original equation? The original equation is sin(...)=1/3. The argument of sine is tan^{-1}(x) + cot^{-1}(1/x). Let me check if that sum is defined for all real x except zero? For x ≠ 0, because 1/x is in the cot^{-1} term. So x can't be zero. But since the equation is defined for all x ≠ 0, and we have solutions x = 3 ± 2√2 and -3 ± 2√2, none of which are zero, so they are all valid.

Therefore, the four solutions are correct. However, wait, maybe I made a mistake in assuming that both positive and negative solutions are valid? Let me check one of the negative solutions numerically.

Take x = - (3 - 2√2) ≈ -0.172. Compute tan^{-1}(x) + cot^{-1}(1/x).

First, tan^{-1}(-0.172) ≈ -0.170 radians (since tan(-0.17) ≈ -0.17). Then 1/x is 1/(-0.172) ≈ -5.814. Then cot^{-1}(-5.814). Let me compute that. Since cot^{-1}(y) is the angle whose cotangent is y. So cot(θ) = -5.814. Which would be in the second quadrant. Let me compute tan(θ) = 1/(-5.814) ≈ -0.172. So θ is π - tan^{-1}(0.172) ≈ π - 0.170 ≈ 3.011 radians. Therefore, tan^{-1}(x) + cot^{-1}(1/x) ≈ -0.170 + 3.011 ≈ 2.841 radians. Now compute sin(2.841). Let me check sin(2.841). Since π ≈ 3.1416, so 2.841 radians is π - 0.3006 radians. Therefore, sin(π - 0.3006) = sin(0.3006) ≈ 0.295. Now, 1/3 ≈ 0.333. Wait, but 0.295 is not equal to 0.333. Wait, this seems like a discrepancy. Did I do something wrong here?

Wait, hold on. Maybe my approximated calculations are leading me astray. Let me calculate more accurately. Let me compute x = 3 - 2√2. First, compute exact value of 3 - 2√2:

√2 ≈ 1.4142, so 2√2 ≈ 2.8284, so 3 - 2.8284 ≈ 0.1716. Therefore, x ≈ -0.1716. Then 1/x ≈ -5.8284.

Compute tan^{-1}(x) where x ≈ -0.1716. Let me compute arctangent of -0.1716. Let me use calculator-like approach. The value arctan(-0.1716) ≈ -0.17 radians (exact value would need calculator, but let me use calculator here). Alternatively, I can use the exact expressions. Let me recall that tan^{-1}(x) + cot^{-1}(1/x) for x negative. Earlier steps suggested that it equals π - 2 tan^{-1}(a), where a = -x. So for x = -a, with a = 3 - 2√2. Therefore, π - 2 tan^{-1}(a). Let me compute sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3. Therefore, if this is true, then the value should be 1/3. However, my approximate calculation gave me about 0.295 instead of 0.333. There's inconsistency here. Therefore, there's a mistake in my reasoning.

Let me check my calculation. Let me compute sin(2 tan^{-1}(a)) where a = 3 - 2√2. Earlier, we had that 2a/(1 + a²) = 1/3. Let me check that. Let me compute a:

a = 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172.

Compute 2a/(1 + a²):

First, compute a²: (3 - 2√2)^2 = 9 - 12√2 + 8 = 17 - 12√2 ≈ 17 - 16.97 ≈ 0.03. Therefore, 1 + a² ≈ 1.03. Then 2a ≈ 2*(0.172) ≈ 0.344. Therefore, 0.344 / 1.03 ≈ 0.334, which is approximately 1/3 (≈0.333). Therefore, it is correct. Therefore, sin(2 tan^{-1}(a)) = 1/3. Therefore, sin(π - 2 tan^{-1}(a)) = 1/3. Therefore, in my earlier approximate calculation, I must have made an error. Let me recalculate the approximate value.

Let me compute π - 2 tan^{-1}(a), where a ≈ 0.172. Let me compute tan^{-1}(0.172). Let me recall that tan(θ) ≈ θ for small θ. So tan^{-1}(0.172) ≈ 0.172 radians. Therefore, 2 tan^{-1}(a) ≈ 0.344 radians. Then π - 0.344 ≈ 3.1416 - 0.344 ≈ 2.7976 radians. Then sin(2.7976). Let me compute sin(2.7976). Let me note that π ≈ 3.1416, so 2.7976 is π - 0.344. So sin(π - 0.344) = sin(0.344) ≈ 0.337. Which is approximately 1/3 ≈ 0.333. Therefore, my previous approximate calculation had an error in the angle, but now with better estimation, it's about 0.337, which is close to 1/3. Therefore, this is consistent. Therefore, the calculation is correct. Therefore, the negative solutions are valid.

Therefore, all four solutions are indeed valid. However, let me check for x = - (3 + 2√2) ≈ -5.828. Let me compute tan^{-1}(x) + cot^{-1}(1/x). Let me denote a = 3 + 2√2 ≈ 5.828. Then, tan^{-1}(x) = tan^{-1}(-a) = -tan^{-1}(a). cot^{-1}(1/x) = cot^{-1}(-1/a) = π - cot^{-1}(1/a) = π - tan^{-1}(a). Therefore, adding them gives:

- tan^{-1}(a) + π - tan^{-1}(a) = π - 2 tan^{-1}(a). Therefore, sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 2a/(1 + a²). Let me compute 2a/(1 + a²). For a = 3 + 2√2, as before, 1 + a² = 1 + (3 + 2√2)^2 = 1 + 17 + 12√2 = 18 + 12√2. Then 2a = 6 + 4√2. Therefore, ratio is (6 + 4√2)/(18 + 12√2) = same as earlier, which is 1/3. Therefore, sin(π - 2 tan^{-1}(a)) = 1/3. Therefore, correct. Therefore, the negative solutions are valid.

Therefore, the equation has four solutions: $3 + 2\sqrt{2}$, $3 - 2\sqrt{2}$, $-3 - 2\sqrt{2}$, and $-3 + 2\sqrt{2}$. But let me check if these are all distinct. Well, $3 + 2\sqrt{2}$ and $-3 - 2\sqrt{2}$ are distinct, and $3 - 2\sqrt{2}$ and $-3 + 2\sqrt{2}$ are distinct. Therefore, four solutions in total.

But let me check if any of these solutions are duplicates or if there's any mistake in the algebra. Wait, the quadratic equation was $x^2 - 6x + 1 = 0$, which has two solutions. Then, when considering negative x, we have two more solutions. Therefore, total four solutions. So the answer should include all four.

But let me check the original problem statement again. It says "Find all solutions to [...]". It doesn't specify if x is positive or negative, so all four should be included. However, let me check if there is any mistake in my analysis.

Wait, but maybe there is a mistake here. Let me think again. The original equation is sin(A + B) = 1/3, where A = tan^{-1}(x), B = cot^{-1}(1/x). For x > 0, we have B = tan^{-1}(x), so A + B = 2 tan^{-1}(x). For x < 0, we have A + B = π - 2 tan^{-1}(-x). However, when I solved, I found that sin(A + B) = sin(2 tan^{-1}(x)) in both cases? Wait, no. For positive x, sin(A + B) = sin(2 tan^{-1}(x)). For negative x, sin(A + B) = sin(π - 2 tan^{-1}(-x)) = sin(2 tan^{-1}(-x))? Wait, no:

Wait, sin(π - θ) = sin θ. Therefore, if for negative x, sin(A + B) = sin(π - 2 tan^{-1}(a)) where a = -x, which is sin(2 tan^{-1}(a)) = 1/3. Therefore, in terms of x, if a = -x, then sin(2 tan^{-1}(-x)) = 1/3? Wait, no, it's sin(π - 2 tan^{-1}(a)) where a = -x. Wait, let me rephrase:

If x is negative, then a = -x is positive, and:

sin(A + B) = sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3. Therefore, we have the same equation for a as before. Which is equivalent to solving 2a/(1 + a²) = 1/3. Therefore, solving for a gives the same quadratic equation. Therefore, a = 3 ± 2√2. Therefore, since a = -x, x = -a = -3 ∓ 2√2. Therefore, the four solutions are x = 3 + 2√2, x = 3 - 2√2, x = -3 - 2√2, x = -3 + 2√2.

Therefore, these are all four solutions.

But let me check once again with another approach. Let me consider the identity:

tan^{-1}(x) + cot^{-1}(x) = π/2 for x > 0. Wait, but here we have cot^{-1}(1/x), which is tan^{-1}(x) for x > 0. Therefore, tan^{-1}(x) + cot^{-1}(1/x) = tan^{-1}(x) + tan^{-1}(x) = 2 tan^{-1}(x). Therefore, for x > 0, it's 2 tan^{-1}(x). For x < 0, as we saw, it's π - 2 tan^{-1}(-x). However, if I think of tan^{-1}(x) + cot^{-1}(x) = π/2 for x > 0. But here, the argument is tan^{-1}(x) + cot^{-1}(1/x). For x > 0, this is tan^{-1}(x) + tan^{-1}(x) = 2 tan^{-1}(x). For x < 0, it's tan^{-1}(x) + cot^{-1}(1/x) = tan^{-1}(x) + π - tan^{-1}(-x) = π + tan^{-1}(x) - tan^{-1}(-x). But tan^{-1}(-x) = -tan^{-1}(x), so this becomes π + tan^{-1}(x) + tan^{-1}(x) = π + 2 tan^{-1}(x). Wait, but earlier I thought it was π - 2 tan^{-1}(-x). Wait, perhaps I miscalculated before?

Wait, let me do this again. For x < 0, let me denote x = -a where a > 0. Then:

tan^{-1}(x) = tan^{-1}(-a) = -tan^{-1}(a)

cot^{-1}(1/x) = cot^{-1}(-1/a) = π - cot^{-1}(1/a) = π - tan^{-1}(a)

Therefore, tan^{-1}(x) + cot^{-1}(1/x) = -tan^{-1}(a) + π - tan^{-1}(a) = π - 2 tan^{-1}(a)

Therefore, the argument inside sine is π - 2 tan^{-1}(a). Therefore, sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3. Therefore, same equation as before. So, the previous analysis is correct. Therefore, for negative x, solving sin(π - 2 tan^{-1}(a)) = 1/3 leads to the same equation as for positive x. Therefore, the four solutions are valid.

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

But let me check if maybe the original equation can be simplified more directly. Let me think again:

Original equation: sin(tan^{-1}(x) + cot^{-1}(1/x)) = 1/3.

Let me use the identity for sin(A + B) = sinA cosB + cosA sinB.

Let me compute A = tan^{-1}(x), B = cot^{-1}(1/x). Let me compute sinA, cosA, sinB, cosB.

For A = tan^{-1}(x): as before, if x is positive, then sinA = x / sqrt(1 + x²), cosA = 1 / sqrt(1 + x²).

For B = cot^{-1}(1/x). Let me think. Let me denote B = cot^{-1}(1/x). Then cotB = 1/x, which means tanB = x. Therefore, B = tan^{-1}(x) if x > 0, which we already considered. However, for general x, B = tan^{-1}(x) if x > 0, but for x < 0, B is in different quadrant. However, if we use the identity for sin(A + B):

sin(A + B) = sinA cosB + cosA sinB.

But since B = cot^{-1}(1/x), which is the same as tan^{-1}(x) when x > 0. Let me check:

If B = cot^{-1}(1/x), then tanB = x. Therefore, B = tan^{-1}(x) + kπ. But since cot^{-1}(1/x) is between 0 and π, and tan^{-1}(x) is between -π/2 and π/2, so depending on x, we have:

For x > 0: tan^{-1}(x) is between 0 and π/2, so B = tan^{-1}(x).

For x < 0: tan^{-1}(x) is between -π/2 and 0. But cot^{-1}(1/x) is between π/2 and π, since 1/x is negative. Let me check for x < 0:

Let me take x negative. Let me denote x = -a, a > 0. Then cot^{-1}(1/x) = cot^{-1}(-1/a) = π - cot^{-1}(1/a) = π - tan^{-1}(a). So B = π - tan^{-1}(a). Therefore, tanB = tan(π - tan^{-1}(a)) = -tan(tan^{-1}(a)) = -a = x. Which is correct. Therefore, for general x, B = tan^{-1}(x) + π * k, but adjusted for quadrants. However, in terms of sin(A + B), perhaps it's easier to proceed.

But sin(A + B) = sinA cosB + cosA sinB. Let me compute sinB and cosB.

For B = cot^{-1}(1/x). Let me consider right triangle where cotB = 1/x. So cotB = adjacent/opposite = 1/x. Therefore, adjacent = 1, opposite = x, hypotenuse sqrt(1 + x²). Therefore, sinB = opposite / hypotenuse = x / sqrt(1 + x²), cosB = adjacent / hypotenuse = 1 / sqrt(1 + x²). Wait, but this is only if B is in the first quadrant. However, if x is negative, then cotB = 1/x is negative. Therefore, B is in the second quadrant. Therefore, sinB is positive (since sine is positive in second quadrant), and cosB is negative. Therefore:

For B = cot^{-1}(1/x):

If x > 0: B is in first quadrant, so sinB = x / sqrt(1 + x²), cosB = 1 / sqrt(1 + x²).

If x < 0: B is in second quadrant. Then, cotB = 1/x is negative, so adjacent = 1, opposite = x (negative), hypotenuse sqrt(1 + x²). But in second quadrant, adjacent is negative, opposite is positive. Wait, this is confusing. Let me think again.

Alternatively, using cotB = 1/x, so tanB = x. Therefore, B is an angle whose tangent is x. Therefore, for x negative, B is in second quadrant. Therefore, sinB = |x| / sqrt(1 + x²), cosB = -1 / sqrt(1 + x²). Wait, let's do it properly. Let me define B such that cotB = 1/x. Therefore, tanB = x. Let me consider B in the appropriate quadrant. For x positive, B is in first quadrant. For x negative, since cotB = 1/x is negative, B is in second quadrant where cotangent is negative (since cotangent is cosine over sine; in second quadrant, cosine is negative, sine is positive, so cotangent is negative). Therefore, tanB = x is negative, which is consistent with B in second quadrant. So for B in second quadrant:

sinB = |x| / sqrt(1 + x²), cosB = -1 / sqrt(1 + x²). Wait, but tanB = sinB / cosB = (|x| / sqrt(...)) / (-1 / sqrt(...)) = -|x|. But tanB = x, which is negative. Therefore, -|x| = x, which implies that x is negative. Therefore, yes, this works. Therefore, in general:

sinB = |x| / sqrt(1 + x²), cosB = -sign(x) / sqrt(1 + x²). Wait, no:

Wait, for B in second quadrant, cosB is negative, sinB is positive. tanB = x < 0. Therefore, tanB = sinB / cosB = (positive) / (negative) = negative. Which matches. Therefore, sinB = |x| / sqrt(1 + x²) if x is negative? Wait, no. Let me think again. Let me write tanB = x. So if x is negative, and B is in second quadrant, then:

Let me represent B as π - θ, where θ is in first quadrant. Then tanB = tan(π - θ) = - tanθ = x. Therefore, tanθ = -x, since x is negative. Therefore, θ = tan^{-1}(-x), which is positive. Therefore, sinB = sin(π - θ) = sinθ = tanθ / sqrt(1 + tan²θ) = (-x)/sqrt(1 + x²). Wait, tanθ = -x, since x is negative. Wait:

Let me explain again. If x is negative, then tanB = x < 0. Let B = π - θ, where θ = tan^{-1}(|x|). Therefore:

sinB = sin(π - θ) = sinθ = |x| / sqrt(1 + x²) [since θ = tan^{-1}(|x|), so sinθ = |x| / sqrt(1 + x²)].

cosB = cos(π - θ) = -cosθ = -1 / sqrt(1 + x²).

Therefore, for general x:

sinB = |x| / sqrt(1 + x²), cosB = -sign(x) / sqrt(1 + x²). Wait, if x is positive, then B is in first quadrant. So previous analysis would be:

sinB = x / sqrt(1 + x²), cosB = 1 / sqrt(1 + x²). For x negative, sinB = |x| / sqrt(1 + x²) = -x / sqrt(1 + x²), cosB = -1 / sqrt(1 + x²).

Therefore, combining both cases:

sinB = x / sqrt(1 + x²) if x > 0, and sinB = -x / sqrt(1 + x²) if x < 0. But wait, for x < 0, |x| = -x, so sinB = |x| / sqrt(...) = -x / sqrt(...). So sinB = sign(x) * x / sqrt(1 + x²)? Wait, no. Wait:

For x positive: sinB = x / sqrt(1 + x²).

For x negative: sinB = (-x)/sqrt(1 + x²) = |x| / sqrt(...). Therefore, sinB = |x| / sqrt(1 + x²).

But if I write this as sinB = x / sqrt(1 + x²) if x > 0, and sinB = -x / sqrt(1 + x²) if x < 0. Therefore, sinB = (x / |x|) * |x| / sqrt(1 + x²) = x / sqrt(1 + x²) if x ≠ 0. Wait, no. Wait:

Wait, for x < 0: |x| = -x, so sinB = |x| / sqrt(1 + x²) = (-x)/sqrt(1 + x²). Which is the same as -x / sqrt(1 + x²). While for x positive, it's x / sqrt... Therefore, combining both, sinB = x / sqrt(1 + x²) when x > 0, and -x / sqrt(1 + x²) when x < 0. But since x / sqrt(1 + x²) is positive for x > 0, negative for x < 0, which matches.

But regardless, let me go back to the original expression:

sin(A + B) = sinA cosB + cosA sinB.

We have A = tan^{-1}(x), so sinA = x / sqrt(1 + x²), cosA = 1 / sqrt(1 + x²).

For B = cot^{-1}(1/x), we have sinB = x / sqrt(1 + x²) if x > 0, and sinB = -x / sqrt(1 + x²) if x < 0, and cosB = 1 / sqrt(1 + x²) if x > 0, and cosB = -1 / sqrt(1 + x²) if x < 0.

Wait, but for x < 0:

sinB = -x / sqrt(1 + x²), and cosB = -1 / sqrt(1 + x²). Therefore, sin(A + B) = sinA cosB + cosA sinB = [x / sqrt(...)] * [ -1 / sqrt(...) ] + [1 / sqrt(...)] * [ -x / sqrt(...) ] = [ -x / (1 + x²) ] + [ -x / (1 + x²) ] = -2x / (1 + x²).

Wait, but for x < 0, we have sin(A + B) = -2x / (1 + x²). Let me check this.

Wait, let me do it with the expressions. For x < 0:

sinA = x / sqrt(1 + x²) (which is negative)

cosB = -1 / sqrt(1 + x²)

Therefore, sinA * cosB = (x / sqrt(1 + x²)) * (-1 / sqrt(1 + x²)) = -x / (1 + x²)

cosA = 1 / sqrt(1 + x²)

sinB = -x / sqrt(1 + x²)

Therefore, cosA * sinB = (1 / sqrt(1 + x²)) * (-x / sqrt(1 + x²)) = -x / (1 + x²)

Therefore, total sin(A + B) = -x / (1 + x²) + (-x) / (1 + x²) = -2x / (1 + x²)

For x > 0, sinA = x / sqrt(...), cosB = 1 / sqrt(...), sinB = x / sqrt(...), cosA = 1 / sqrt(...)

Therefore, sinA * cosB + cosA * sinB = [x * 1 + 1 * x] / (1 + x²) = 2x / (1 + x²)

Therefore, for x > 0: sin(A + B) = 2x / (1 + x²)

For x < 0: sin(A + B) = -2x / (1 + x²)

Therefore, putting it all together:

sin(tan^{-1}(x) + cot^{-1}(1/x)) = { 2x / (1 + x²) if x > 0, -2x / (1 + x²) if x < 0 }

Therefore, the original equation:

sin(...) = 1/3 is equivalent to:

For x > 0: 2x / (1 + x²) = 1/3 => 6x = 1 + x² => x² - 6x + 1 = 0 => x = [6 ± sqrt(36 - 4)] / 2 = [6 ± sqrt(32)] / 2 = 3 ± 2√2. Both positive.

For x < 0: -2x / (1 + x²) = 1/3 => -2x = (1 + x²)/3 => -6x = 1 + x² => x² + 6x + 1 = 0 => x = [-6 ± sqrt(36 - 4)] / 2 = [-6 ± sqrt(32)] / 2 = [-6 ± 4√2]/2 = -3 ± 2√2.

But since x < 0, we check which solutions are negative:

-3 + 2√2 ≈ -3 + 2.828 ≈ -0.172 < 0.

-3 - 2√2 ≈ -5.828 < 0.

Therefore, for x < 0, solutions are x = -3 + 2√2 and x = -3 - 2√2.

Therefore, combining both cases, solutions are:

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

Which matches the previous result. Therefore, this confirms that there are four solutions.

Therefore, the answer is all four values. Therefore, boxed answers:

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

But the problem says "Enter all the solutions, separated by commas." So I need to present them as a list within a single box, separated by commas. Wait, the instructions say "put your final answer within \boxed{}". But usually, for multiple answers, we can write multiple boxed answers or separate them in one box. However, standard practice for multiple solutions is to list them all in one box separated by commas. For example:

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

But maybe as a single box with commas. Let me check previous examples. For example, if there are multiple solutions, they are often listed within a single \boxed{} separated by commas. For example, \boxed{-3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{3 + 2\sqrt{2}}. But instructions say "Enter all the solutions, separated by commas." So likely as a single box with commas.

Alternatively, sometimes problems expect solutions in a specific order or sorted. Let me check approximate values:

3 + 2√2 ≈ 5.828

3 - 2√2 ≈ 0.172

-3 + 2√2 ≈ -0.172

-3 - 2√2 ≈ -5.828

So sorted order: -5.828, -0.172, 0.172, 5.828. So solutions are -3 - 2√2, -3 + 2√2, 3 - 2√2, 3 + 2√2.

Therefore, final answer is these four values. So boxed as:

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

Alternatively, if the system allows multiple boxes, but likely the user wants them in a single box separated by commas. Since the original instruction says "put your final answer within \boxed{}", maybe they expect multiple boxed answers? However, standard practice on some platforms is to have a single box. Given the ambiguity, but following examples from other problems, sometimes multiple boxed answers are acceptable, but if only one box, then comma-separated inside.

Looking at the problem statement again: "Enter all the solutions, separated by commas." So likely they want comma-separated inside a single box. So:

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

But I need to check if this is acceptable. For example, in some contexts, they might expect ordered or simplified, but given that the problem didn't specify, it's safe to list them all.

Therefore, the final answer is the four solutions as above.

**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 if only one box is allowed:

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

But different platforms may vary. Since the user instruction says "put your final answer within \boxed{}", I think they expect multiple boxed answers. However, checking standard practice here, sometimes multiple boxed answers are used for multiple solutions. However, given the original instructions say "Enter all the solutions, separated by commas." and "put your final answer within \boxed{}", I think the expected format is all solutions in one box separated by commas. Therefore:

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

But to adhere to the instruction of one box, I think the correct way is:

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

However, I need to check if the system accepts multiple boxes or expects a single box. Since I can't be sure, but following the initial problem's examples, I think the correct way is multiple boxed answers. However, to conform to standard practice here, I'll put all in one box separated by commas.

**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 system might require only one box, I'll check again. In many online platforms, when multiple solutions are expected, they are put in one box separated by commas. For example, on AoPS or similar, the format is multiple boxed answers. However, given the original problem says "put your final answer within \boxed{}", I think they want one box with all solutions. Hence:

**Final Answer**
\boxed{-3 - 2\sqrt{2}, -3 + 2\sqrt{2}, 3 - 2\sqrt{2}, 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 argument of the sine function:
$$
\tan^{-1}(x) + \cot^{-1}\left( \frac{1}{x} \right).
$$

---

### Step 1: Simplify the Inverse Trigonometric Sum

We use the identity:
$$
\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}, \quad \text{for } x > 0.
$$

Also, recall that:
$$
\cot^{-1}\left( \frac{1}{x} \right) = \tan^{-1}(x), \quad \text{for } x > 0.
$$

Thus, when $ x > 0 $, the expression becomes:
$$
\tan^{-1}(x) + \tan^{-1}(x) = 2 \tan^{-1}(x).
$$

So we now rewrite the original equation for $ x > 0 $ as:
$$
\sin\left( 2 \tan^{-1}(x) \right) = \frac{1}{3}.
$$

Let $ \theta = \tan^{-1}(x) $, so $ \sin(2\theta) = \frac{1}{3} $.

---

### Step 2: Express in Terms of $ x $

Using the double angle identity for sine:
$$
\sin(2\theta) = 2 \sin(\theta) \cos(\theta).
$$

Given that $ \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}} $ and $ \cos(\tan^{-1}(x)) = \frac{1}{\sqrt{1 + x^2}} $, we compute:
$$
\sin(2 \tan^{-1}(x)) = \frac{2x}{1 + x^2}.
$$

Thus, the equation becomes:
$$
\frac{2x}{1 + x^2} = \frac{1}{3}.
$$

---

### Step 3: Solve the Resulting Quadratic Equation

Cross-multiplying:
$$
6x = 1 + x^2 \Rightarrow x^2 - 6x + 1 = 0.
$$

Solving:
$$
x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(1)}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}.
$$

So the two **positive** solutions are:
$$
x = 3 + 2\sqrt{2}, \quad x = 3 - 2\sqrt{2}.
$$

---

### Step 4: Consider the Negative Case

When $ x < 0 $, let $ x = -a $, where $ a > 0 $. Then:

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

Hence:
$$
\tan^{-1}(x) + \cot^{-1}(1/x) = -\tan^{-1}(a) + \pi - \tan^{-1}(a) = \pi - 2\tan^{-1}(a).
$$

Now use the identity $ \sin(\pi - \theta) = \sin(\theta) $, so:
$$
\sin(\pi - 2\tan^{-1}(a)) = \sin(2\tan^{-1}(a)) = \frac{1}{3}.
$$

This leads to the **same equation** as before:
$$
\frac{2a}{1 + a^2} = \frac{1}{3}.
$$

This results in the same quadratic:
$$
a^2 - 6a + 1 = 0 \Rightarrow a = 3 \pm 2\sqrt{2}.
$$

Since $ x = -a $, the **negative** solutions are:
$$
x = -3 - 2\sqrt{2}, \quad x = -3 + 2\sqrt{2}.
$$

---

### Step 5: Final Answer

All **four** real solutions are:

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

---

### Final Answer

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

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

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

First, I remember that inverse trigonometric functions can sometimes be tricky, but maybe there's a relationship between $\tan^{-1}(x)$ and $\cot^{-1}(1/x)$. Let me think... I recall that $\cot^{-1}(y)$ is equal to $\tan^{-1}(1/y)$ when $y > 0$. Is that right? Let me verify. If I consider $\theta = \cot^{-1}(y)$, then $\cot(\theta) = y$, which means $\tan(\theta) = 1/y$, so $\theta = \tan^{-1}(1/y)$. Yeah, that seems correct. So, if $y = 1/x$, then $\cot^{-1}(1/x) = \tan^{-1}(x)$. Wait, but only if $1/x > 0$, which would mean $x > 0$. Hmm, but what if $x$ is negative? Then $1/x$ would be negative, and $\cot^{-1}(1/x)$ would be in a different quadrant? Let me think about that.

But before getting bogged down with the domain issues, maybe I can just assume $x > 0$ first and see where that leads me. Maybe there are solutions for negative $x$ as well? Let's check.

So, if $x > 0$, then $\cot^{-1}(1/x) = \tan^{-1}(x)$. Therefore, the argument inside the sine function becomes $\tan^{-1}(x) + \tan^{-1}(x) = 2\tan^{-1}(x)$. Therefore, the equation simplifies to:

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

Okay, that seems manageable. Let me work with that. Let me denote $\theta = \tan^{-1}(x)$, so that $\theta$ is in the first quadrant because $x > 0$. Then the equation becomes:

$$
\sin(2\theta) = \frac{1}{3}.
$$

Using the double-angle identity, $\sin(2\theta) = 2\sin\theta \cos\theta$. Also, since $\theta = \tan^{-1}(x)$, we can think of a right triangle where the opposite side is $x$ and adjacent is 1, so hypotenuse is $\sqrt{1 + x^2}$. Therefore, $\sin\theta = \frac{x}{\sqrt{1 + x^2}}$ and $\cos\theta = \frac{1}{\sqrt{1 + x^2}}$. Therefore:

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

Therefore, the equation becomes:

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

Cross-multiplying:

$$
6x = 1 + x^2
$$
$$
x^2 - 6x + 1 = 0.
$$

Solving this quadratic equation:

$$
x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 1}}{2} = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}.
$$

So, $x = 3 + 2\sqrt{2}$ or $x = 3 - 2\sqrt{2}$. Let me check if these are positive. Since $\sqrt{2} \approx 1.414, so 2√2 ≈ 2.828. Therefore, 3 - 2.828 ≈ 0.172, which is positive. So both solutions are positive, so they are valid under our initial assumption that $x > 0$.

But wait, what if $x$ is negative? Let me check if there could be solutions for negative $x$. Let me consider $x < 0$. Then, $\cot^{-1}(1/x)$... Let me recall that $\cot^{-1}(y)$ is defined for all real numbers, and it returns values in (0, π). If $y$ is negative, $\cot^{-1}(y)$ would be in (π/2, π). However, $\tan^{-1}(x)$ for $x < 0$ would be in (-π/2, 0).

But if I try to relate $\tan^{-1}(x)$ and $\cot^{-1}(1/x)$, when $x < 0$, is there a relationship? Let me think. Let me take $x = -a$ where $a > 0$. Then $\cot^{-1}(1/x) = \cot^{-1}(-1/a)$. But $\cot^{-1}(-1/a)$ is equal to π - $\cot^{-1}(1/a)$. Because cotangent is negative in the second quadrant. So, $\cot^{-1}(-y) = π - \cot^{-1}(y)$ for $y > 0$. Therefore, $\cot^{-1}(-1/a) = π - \cot^{-1}(1/a)$.

On the other hand, $\tan^{-1}(x) = \tan^{-1}(-a) = -\tan^{-1}(a)$.

Therefore, the argument inside the sine function is:

$$
\tan^{-1}(x) + \cot^{-1}(1/x) = -\tan^{-1}(a) + π - \cot^{-1}(1/a).
$$

But earlier, for positive $a$, we had $\cot^{-1}(1/a) = \tan^{-1}(a)$. So substituting that in:

$$
- \tan^{-1}(a) + π - \tan^{-1}(a) = π - 2 \tan^{-1}(a).
$$

So, the equation becomes:

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

But $\sin(π - θ) = \sin θ$, so this simplifies to:

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

Wait, that's the same equation as before! So even when $x$ is negative, we end up with the same equation as when $x$ is positive. Therefore, solving $\sin(2θ) = 1/3$ would give us solutions for both positive and negative $x$. However, we need to check if there are additional constraints.

But let me think again. If $x$ is negative, then $a = -x > 0$, so solving for $a$ gives the same quadratic equation as before. So:

$$
\frac{2a}{1 + a^2} = \frac{1}{3} \implies a^2 - 6a + 1 = 0 \implies a = 3 \pm 2\sqrt{2}.
$$

Since $a > 0$, both solutions are valid. Therefore, $x = -a$ would be $x = - (3 + 2\sqrt{2})$ and $x = - (3 - 2\sqrt{2})$.

Wait a second, but we need to check if these solutions actually satisfy the original equation. Because sometimes when dealing with inverse trigonometric functions, especially with different quadrants, we might get extraneous solutions.

But let's think again. For the case where $x$ is negative, we had:

$$
\sin\left( π - 2 \tan^{-1}(a) \right) = \sin(2 \tan^{-1}(a)) = 1/3.
$$

But since $\sin(π - θ) = \sin θ$, and we already have that equation. Therefore, the equation for $a$ is the same as for positive $x$, so the solutions for $a$ are $3 \pm 2\sqrt{2}$, so $x = - (3 \pm 2\sqrt{2})$.

But wait, but we have to check if these values are valid. Let me check with $x = 3 + 2\sqrt{2}$, which is positive. Let me compute the original expression. But maybe there's something else. Let me check if the argument inside the sine function is within the correct range.

For positive $x$, the argument is $2 \tan^{-1}(x)$. Since $x > 0$, $\tan^{-1}(x)$ is between 0 and π/2, so doubling it would be between 0 and π. Therefore, the sine of that is between 0 and 1, which is okay.

For negative $x$, the argument was $π - 2 \tan^{-1}(a)$, where $a > 0$. Since $a > 0$, $\tan^{-1}(a)$ is between 0 and π/2, so $2 \tan^{-1}(a)$ is between 0 and π, so π - 2 tan^{-1}(a) is between 0 and π. Therefore, the sine is also between 0 and 1, which is okay. Therefore, both positive and negative solutions are valid.

Therefore, the solutions are $x = 3 + 2\sqrt{2}$, $x = 3 - 2\sqrt{2}$, $x = - (3 + 2\sqrt{2})$, and $x = - (3 - 2\sqrt{2})$?

Wait, but hold on. Let me check if there are only two solutions or four solutions. Wait, the equation $\sin(\theta) = 1/3$ has two solutions in [0, π], which is the range we have for the argument of the sine function (either $2 \tan^{-1}(x)$ or $π - 2 \tan^{-1}(a)$). So, if we consider the general solution for $\sin(\theta) = 1/3$, we have $\theta = \arcsin(1/3)$ or $\theta = π - \arcsin(1/3)$.

But in our case, for positive $x$, the argument is $2 \tan^{-1}(x)$, which is in (0, π). Therefore, solving $2 \tan^{-1}(x) = \arcsin(1/3)$ or $2 \tan^{-1}(x) = π - \arcsin(1/3)$. Wait, but earlier when I set up the equation, I had $\sin(2 \tan^{-1}(x)) = 1/3$, which would correspond to $2 \tan^{-1}(x)$ being in (0, π). Therefore, it's possible that there are two solutions for $x$ in positive numbers. Similarly for negative $x$, but since we ended up with same equation, but $x$ negative. However, in our earlier step, when solving the quadratic, we found two solutions for $a$, which led to two solutions for $x$. So maybe there are only two solutions? Wait, maybe I confused something here.

Wait, let me backtrack. When I considered $x > 0$, I had the equation $\frac{2x}{1 + x^2} = 1/3$, which led to $x^2 - 6x + 1 = 0$, whose solutions are $x = 3 \pm 2\sqrt{2}$. Both positive, so both valid for $x > 0$. Then for $x < 0$, we found that the equation reduces to the same $\frac{2a}{1 + a^2} = 1/3$, with $a = -x > 0$, which gives $a = 3 \pm 2\sqrt{2}$, so $x = -a$ is $x = -3 \pm 2\sqrt{2}$. However, $-3 - 2\sqrt{2}$ is negative, and $-3 + 2\sqrt{2}$ is approximately -3 + 2.828 ≈ -0.172, which is also negative.

Therefore, in total, there are four solutions: $3 + 2\sqrt{2}$, $3 - 2\sqrt{2}$, $-3 - 2\sqrt{2}$, and $-3 + 2\sqrt{2}$. Wait, but hold on, maybe not? Let me check.

Wait, if we have four solutions, but perhaps some of them are extraneous? Let me check with specific values.

Let me first check $x = 3 + 2\sqrt{2}$. Let's compute $\tan^{-1}(x) + \cot^{-1}(1/x)$. Since $x > 0$, $\cot^{-1}(1/x) = \tan^{-1}(x)$. Therefore, the sum is $2 \tan^{-1}(x)$. Then $\sin(2 \tan^{-1}(x))$. As we had earlier, this equals $2x/(1 + x^2)$. Let's compute that with $x = 3 + 2\sqrt{2}$:

First compute $x = 3 + 2\sqrt{2}$. Then $1 + x^2 = 1 + (3 + 2\sqrt{2})^2$. Let me compute that:

$ (3 + 2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} $. So $1 + x^2 = 18 + 12\sqrt{2}$. Then $2x = 2*(3 + 2\sqrt{2}) = 6 + 4\sqrt{2}$. Therefore:

$ \frac{2x}{1 + x^2} = \frac{6 + 4\sqrt{2}}{18 + 12\sqrt{2}} $. Let's rationalize the denominator. Multiply numerator and denominator by the conjugate of the denominator's denominator? Wait, denominator is 18 + 12√2. Alternatively, factor numerator and denominator:

Numerator: 6 + 4√2 = 2*(3 + 2√2)

Denominator: 18 + 12√2 = 6*(3 + 2√2)

Therefore, the ratio is 2*(3 + 2√2)/[6*(3 + 2√2)] = 2/6 = 1/3. Which matches. So yes, that's correct.

Similarly for $x = 3 - 2\sqrt{2}$. Let's check that:

x = 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172. Then 1 + x^2:

x² = (3 - 2√2)^2 = 9 - 12√2 + 8 = 17 - 12√2. Therefore, 1 + x² = 18 - 12√2. Then 2x = 6 - 4√2. Therefore:

(6 - 4√2)/(18 - 12√2). Again, factor numerator and denominator:

Numerator: 2*(3 - 2√2)

Denominator: 6*(3 - 2√2)

So ratio is 2/6 = 1/3. Correct.

Now check the negative solutions. Let's take x = -3 - 2√2. Let me compute $\tan^{-1}(x) + \cot^{-1}(1/x)$. Since x is negative, let me use the previous analysis. Let me denote a = -x = 3 + 2√2. Then:

$\tan^{-1}(x) + \cot^{-1}(1/x) = -\tan^{-1}(a) + π - \cot^{-1}(1/a)$. But since a > 0, $\cot^{-1}(1/a) = \tan^{-1}(a)$. Therefore:

- tan^{-1}(a) + π - tan^{-1}(a) = π - 2 tan^{-1}(a). Then sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3, as before. So same result. Therefore, the value is 1/3. Therefore, x = -a = -3 - 2√2 is valid.

Similarly for x = -3 + 2√2. Let me check that. Let a = -x = 3 - 2√2. Then the same steps apply. So sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3. Which again, for a = 3 - 2√2, we already saw that 2a/(1 + a²) = 1/3. Therefore, this also holds. Therefore, all four solutions are valid.

But let me check if all four solutions are distinct. Let me compute approximate numerical values:

3 + 2√2 ≈ 3 + 2.828 ≈ 5.828

3 - 2√2 ≈ 0.172

-3 - 2√2 ≈ -5.828

-3 + 2√2 ≈ -0.172

Yes, all four are distinct. Therefore, there are four solutions. However, I need to check if the original equation allows all four. Wait, but maybe there's a mistake here. Let me check for x negative. Let me pick x = -0.172, which is - (3 - 2√2). Let me compute the original expression:

sin( tan^{-1}(x) + cot^{-1}(1/x) )

First compute tan^{-1}(x). Since x is negative, tan^{-1}(x) is in (-π/2, 0). Let me compute tan^{-1}(-0.172). Let me denote that as θ. Then cot^{-1}(1/x) = cot^{-1}(1/(-0.172)) = cot^{-1}(-5.828). But cot^{-1}(-5.828) is equal to π - cot^{-1}(5.828). Let me compute cot^{-1}(5.828). Since cot^{-1}(y) is equal to tan^{-1}(1/y), so cot^{-1}(5.828) is tan^{-1}(1/5.828) ≈ tan^{-1}(0.172) ≈ θ' where tan θ' ≈ 0.172, which is approximately θ, since tan(θ) = -0.172. But θ is negative, so θ' is positive. Therefore, cot^{-1}(-5.828) = π - θ'.

So tan^{-1}(x) + cot^{-1}(1/x) = θ + π - θ' = π + θ - θ'. But θ = tan^{-1}(x) = tan^{-1}(-0.172) ≈ -θ', since tan(-θ') = -tan(theta') ≈ -0.172. Therefore, θ ≈ -θ', so θ - θ' ≈ -θ' - θ' = -2θ'. Therefore, the total expression is π - 2θ', which is π - 2 tan^{-1}(a), with a = 3 - 2√2 ≈ 0.172. Therefore, sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3. So that works.

Similarly, for the other negative value. Therefore, all four solutions are valid. Therefore, the answer should have four solutions: 3 + 2√2, 3 - 2√2, -3 - 2√2, -3 + 2√2.

But let me check if there's any other constraints. For example, is there any domain restrictions on the original equation? The original equation is sin(...)=1/3. The argument of sine is tan^{-1}(x) + cot^{-1}(1/x). Let me check if that sum is defined for all real x except zero? For x ≠ 0, because 1/x is in the cot^{-1} term. So x can't be zero. But since the equation is defined for all x ≠ 0, and we have solutions x = 3 ± 2√2 and -3 ± 2√2, none of which are zero, so they are all valid.

Therefore, the four solutions are correct. However, wait, maybe I made a mistake in assuming that both positive and negative solutions are valid? Let me check one of the negative solutions numerically.

Take x = - (3 - 2√2) ≈ -0.172. Compute tan^{-1}(x) + cot^{-1}(1/x).

First, tan^{-1}(-0.172) ≈ -0.170 radians (since tan(-0.17) ≈ -0.17). Then 1/x is 1/(-0.172) ≈ -5.814. Then cot^{-1}(-5.814). Let me compute that. Since cot^{-1}(y) is the angle whose cotangent is y. So cot(θ) = -5.814. Which would be in the second quadrant. Let me compute tan(θ) = 1/(-5.814) ≈ -0.172. So θ is π - tan^{-1}(0.172) ≈ π - 0.170 ≈ 3.011 radians. Therefore, tan^{-1}(x) + cot^{-1}(1/x) ≈ -0.170 + 3.011 ≈ 2.841 radians. Now compute sin(2.841). Let me check sin(2.841). Since π ≈ 3.1416, so 2.841 radians is π - 0.3006 radians. Therefore, sin(π - 0.3006) = sin(0.3006) ≈ 0.295. Now, 1/3 ≈ 0.333. Wait, but 0.295 is not equal to 0.333. Wait, this seems like a discrepancy. Did I do something wrong here?

Wait, hold on. Maybe my approximated calculations are leading me astray. Let me calculate more accurately. Let me compute x = 3 - 2√2. First, compute exact value of 3 - 2√2:

√2 ≈ 1.4142, so 2√2 ≈ 2.8284, so 3 - 2.8284 ≈ 0.1716. Therefore, x ≈ -0.1716. Then 1/x ≈ -5.8284.

Compute tan^{-1}(x) where x ≈ -0.1716. Let me compute arctangent of -0.1716. Let me use calculator-like approach. The value arctan(-0.1716) ≈ -0.17 radians (exact value would need calculator, but let me use calculator here). Alternatively, I can use the exact expressions. Let me recall that tan^{-1}(x) + cot^{-1}(1/x) for x negative. Earlier steps suggested that it equals π - 2 tan^{-1}(a), where a = -x. So for x = -a, with a = 3 - 2√2. Therefore, π - 2 tan^{-1}(a). Let me compute sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3. Therefore, if this is true, then the value should be 1/3. However, my approximate calculation gave me about 0.295 instead of 0.333. There's inconsistency here. Therefore, there's a mistake in my reasoning.

Let me check my calculation. Let me compute sin(2 tan^{-1}(a)) where a = 3 - 2√2. Earlier, we had that 2a/(1 + a²) = 1/3. Let me check that. Let me compute a:

a = 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172.

Compute 2a/(1 + a²):

First, compute a²: (3 - 2√2)^2 = 9 - 12√2 + 8 = 17 - 12√2 ≈ 17 - 16.97 ≈ 0.03. Therefore, 1 + a² ≈ 1.03. Then 2a ≈ 2*(0.172) ≈ 0.344. Therefore, 0.344 / 1.03 ≈ 0.334, which is approximately 1/3 (≈0.333). Therefore, it is correct. Therefore, sin(2 tan^{-1}(a)) = 1/3. Therefore, sin(π - 2 tan^{-1}(a)) = 1/3. Therefore, in my earlier approximate calculation, I must have made an error. Let me recalculate the approximate value.

Let me compute π - 2 tan^{-1}(a), where a ≈ 0.172. Let me compute tan^{-1}(0.172). Let me recall that tan(θ) ≈ θ for small θ. So tan^{-1}(0.172) ≈ 0.172 radians. Therefore, 2 tan^{-1}(a) ≈ 0.344 radians. Then π - 0.344 ≈ 3.1416 - 0.344 ≈ 2.7976 radians. Then sin(2.7976). Let me compute sin(2.7976). Let me note that π ≈ 3.1416, so 2.7976 is π - 0.344. So sin(π - 0.344) = sin(0.344) ≈ 0.337. Which is approximately 1/3 ≈ 0.333. Therefore, my previous approximate calculation had an error in the angle, but now with better estimation, it's about 0.337, which is close to 1/3. Therefore, this is consistent. Therefore, the calculation is correct. Therefore, the negative solutions are valid.

Therefore, all four solutions are indeed valid. However, let me check for x = - (3 + 2√2) ≈ -5.828. Let me compute tan^{-1}(x) + cot^{-1}(1/x). Let me denote a = 3 + 2√2 ≈ 5.828. Then, tan^{-1}(x) = tan^{-1}(-a) = -tan^{-1}(a). cot^{-1}(1/x) = cot^{-1}(-1/a) = π - cot^{-1}(1/a) = π - tan^{-1}(a). Therefore, adding them gives:

- tan^{-1}(a) + π - tan^{-1}(a) = π - 2 tan^{-1}(a). Therefore, sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 2a/(1 + a²). Let me compute 2a/(1 + a²). For a = 3 + 2√2, as before, 1 + a² = 1 + (3 + 2√2)^2 = 1 + 17 + 12√2 = 18 + 12√2. Then 2a = 6 + 4√2. Therefore, ratio is (6 + 4√2)/(18 + 12√2) = same as earlier, which is 1/3. Therefore, sin(π - 2 tan^{-1}(a)) = 1/3. Therefore, correct. Therefore, the negative solutions are valid.

Therefore, the equation has four solutions: $3 + 2\sqrt{2}$, $3 - 2\sqrt{2}$, $-3 - 2\sqrt{2}$, and $-3 + 2\sqrt{2}$. But let me check if these are all distinct. Well, $3 + 2\sqrt{2}$ and $-3 - 2\sqrt{2}$ are distinct, and $3 - 2\sqrt{2}$ and $-3 + 2\sqrt{2}$ are distinct. Therefore, four solutions in total.

But let me check if any of these solutions are duplicates or if there's any mistake in the algebra. Wait, the quadratic equation was $x^2 - 6x + 1 = 0$, which has two solutions. Then, when considering negative x, we have two more solutions. Therefore, total four solutions. So the answer should include all four.

But let me check the original problem statement again. It says "Find all solutions to [...]". It doesn't specify if x is positive or negative, so all four should be included. However, let me check if there is any mistake in my analysis.

Wait, but maybe there is a mistake here. Let me think again. The original equation is sin(A + B) = 1/3, where A = tan^{-1}(x), B = cot^{-1}(1/x). For x > 0, we have B = tan^{-1}(x), so A + B = 2 tan^{-1}(x). For x < 0, we have A + B = π - 2 tan^{-1}(-x). However, when I solved, I found that sin(A + B) = sin(2 tan^{-1}(x)) in both cases? Wait, no. For positive x, sin(A + B) = sin(2 tan^{-1}(x)). For negative x, sin(A + B) = sin(π - 2 tan^{-1}(-x)) = sin(2 tan^{-1}(-x))? Wait, no:

Wait, sin(π - θ) = sin θ. Therefore, if for negative x, sin(A + B) = sin(π - 2 tan^{-1}(a)) where a = -x, which is sin(2 tan^{-1}(a)) = 1/3. Therefore, in terms of x, if a = -x, then sin(2 tan^{-1}(-x)) = 1/3? Wait, no, it's sin(π - 2 tan^{-1}(a)) where a = -x. Wait, let me rephrase:

If x is negative, then a = -x is positive, and:

sin(A + B) = sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3. Therefore, we have the same equation for a as before. Which is equivalent to solving 2a/(1 + a²) = 1/3. Therefore, solving for a gives the same quadratic equation. Therefore, a = 3 ± 2√2. Therefore, since a = -x, x = -a = -3 ∓ 2√2. Therefore, the four solutions are x = 3 + 2√2, x = 3 - 2√2, x = -3 - 2√2, x = -3 + 2√2.

Therefore, these are all four solutions.

But let me check once again with another approach. Let me consider the identity:

tan^{-1}(x) + cot^{-1}(x) = π/2 for x > 0. Wait, but here we have cot^{-1}(1/x), which is tan^{-1}(x) for x > 0. Therefore, tan^{-1}(x) + cot^{-1}(1/x) = tan^{-1}(x) + tan^{-1}(x) = 2 tan^{-1}(x). Therefore, for x > 0, it's 2 tan^{-1}(x). For x < 0, as we saw, it's π - 2 tan^{-1}(-x). However, if I think of tan^{-1}(x) + cot^{-1}(x) = π/2 for x > 0. But here, the argument is tan^{-1}(x) + cot^{-1}(1/x). For x > 0, this is tan^{-1}(x) + tan^{-1}(x) = 2 tan^{-1}(x). For x < 0, it's tan^{-1}(x) + cot^{-1}(1/x) = tan^{-1}(x) + π - tan^{-1}(-x) = π + tan^{-1}(x) - tan^{-1}(-x). But tan^{-1}(-x) = -tan^{-1}(x), so this becomes π + tan^{-1}(x) + tan^{-1}(x) = π + 2 tan^{-1}(x). Wait, but earlier I thought it was π - 2 tan^{-1}(-x). Wait, perhaps I miscalculated before?

Wait, let me do this again. For x < 0, let me denote x = -a where a > 0. Then:

tan^{-1}(x) = tan^{-1}(-a) = -tan^{-1}(a)

cot^{-1}(1/x) = cot^{-1}(-1/a) = π - cot^{-1}(1/a) = π - tan^{-1}(a)

Therefore, tan^{-1}(x) + cot^{-1}(1/x) = -tan^{-1}(a) + π - tan^{-1}(a) = π - 2 tan^{-1}(a)

Therefore, the argument inside sine is π - 2 tan^{-1}(a). Therefore, sin(π - 2 tan^{-1}(a)) = sin(2 tan^{-1}(a)) = 1/3. Therefore, same equation as before. So, the previous analysis is correct. Therefore, for negative x, solving sin(π - 2 tan^{-1}(a)) = 1/3 leads to the same equation as for positive x. Therefore, the four solutions are valid.

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

But let me check if maybe the original equation can be simplified more directly. Let me think again:

Original equation: sin(tan^{-1}(x) + cot^{-1}(1/x)) = 1/3.

Let me use the identity for sin(A + B) = sinA cosB + cosA sinB.

Let me compute A = tan^{-1}(x), B = cot^{-1}(1/x). Let me compute sinA, cosA, sinB, cosB.

For A = tan^{-1}(x): as before, if x is positive, then sinA = x / sqrt(1 + x²), cosA = 1 / sqrt(1 + x²).

For B = cot^{-1}(1/x). Let me think. Let me denote B = cot^{-1}(1/x). Then cotB = 1/x, which means tanB = x. Therefore, B = tan^{-1}(x) if x > 0, which we already considered. However, for general x, B = tan^{-1}(x) if x > 0, but for x < 0, B is in different quadrant. However, if we use the identity for sin(A + B):

sin(A + B) = sinA cosB + cosA sinB.

But since B = cot^{-1}(1/x), which is the same as tan^{-1}(x) when x > 0. Let me check:

If B = cot^{-1}(1/x), then tanB = x. Therefore, B = tan^{-1}(x) + kπ. But since cot^{-1}(1/x) is between 0 and π, and tan^{-1}(x) is between -π/2 and π/2, so depending on x, we have:

For x > 0: tan^{-1}(x) is between 0 and π/2, so B = tan^{-1}(x).

For x < 0: tan^{-1}(x) is between -π/2 and 0. But cot^{-1}(1/x) is between π/2 and π, since 1/x is negative. Let me check for x < 0:

Let me take x negative. Let me denote x = -a, a > 0. Then cot^{-1}(1/x) = cot^{-1}(-1/a) = π - cot^{-1}(1/a) = π - tan^{-1}(a). So B = π - tan^{-1}(a). Therefore, tanB = tan(π - tan^{-1}(a)) = -tan(tan^{-1}(a)) = -a = x. Which is correct. Therefore, for general x, B = tan^{-1}(x) + π * k, but adjusted for quadrants. However, in terms of sin(A + B), perhaps it's easier to proceed.

But sin(A + B) = sinA cosB + cosA sinB. Let me compute sinB and cosB.

For B = cot^{-1}(1/x). Let me consider right triangle where cotB = 1/x. So cotB = adjacent/opposite = 1/x. Therefore, adjacent = 1, opposite = x, hypotenuse sqrt(1 + x²). Therefore, sinB = opposite / hypotenuse = x / sqrt(1 + x²), cosB = adjacent / hypotenuse = 1 / sqrt(1 + x²). Wait, but this is only if B is in the first quadrant. However, if x is negative, then cotB = 1/x is negative. Therefore, B is in the second quadrant. Therefore, sinB is positive (since sine is positive in second quadrant), and cosB is negative. Therefore:

For B = cot^{-1}(1/x):

If x > 0: B is in first quadrant, so sinB = x / sqrt(1 + x²), cosB = 1 / sqrt(1 + x²).

If x < 0: B is in second quadrant. Then, cotB = 1/x is negative, so adjacent = 1, opposite = x (negative), hypotenuse sqrt(1 + x²). But in second quadrant, adjacent is negative, opposite is positive. Wait, this is confusing. Let me think again.

Alternatively, using cotB = 1/x, so tanB = x. Therefore, B is an angle whose tangent is x. Therefore, for x negative, B is in second quadrant. Therefore, sinB = |x| / sqrt(1 + x²), cosB = -1 / sqrt(1 + x²). Wait, let's do it properly. Let me define B such that cotB = 1/x. Therefore, tanB = x. Let me consider B in the appropriate quadrant. For x positive, B is in first quadrant. For x negative, since cotB = 1/x is negative, B is in second quadrant where cotangent is negative (since cotangent is cosine over sine; in second quadrant, cosine is negative, sine is positive, so cotangent is negative). Therefore, tanB = x is negative, which is consistent with B in second quadrant. So for B in second quadrant:

sinB = |x| / sqrt(1 + x²), cosB = -1 / sqrt(1 + x²). Wait, but tanB = sinB / cosB = (|x| / sqrt(...)) / (-1 / sqrt(...)) = -|x|. But tanB = x, which is negative. Therefore, -|x| = x, which implies that x is negative. Therefore, yes, this works. Therefore, in general:

sinB = |x| / sqrt(1 + x²), cosB = -sign(x) / sqrt(1 + x²). Wait, no:

Wait, for B in second quadrant, cosB is negative, sinB is positive. tanB = x < 0. Therefore, tanB = sinB / cosB = (positive) / (negative) = negative. Which matches. Therefore, sinB = |x| / sqrt(1 + x²) if x is negative? Wait, no. Let me think again. Let me write tanB = x. So if x is negative, and B is in second quadrant, then:

Let me represent B as π - θ, where θ is in first quadrant. Then tanB = tan(π - θ) = - tanθ = x. Therefore, tanθ = -x, since x is negative. Therefore, θ = tan^{-1}(-x), which is positive. Therefore, sinB = sin(π - θ) = sinθ = tanθ / sqrt(1 + tan²θ) = (-x)/sqrt(1 + x²). Wait, tanθ = -x, since x is negative. Wait:

Let me explain again. If x is negative, then tanB = x < 0. Let B = π - θ, where θ = tan^{-1}(|x|). Therefore:

sinB = sin(π - θ) = sinθ = |x| / sqrt(1 + x²) [since θ = tan^{-1}(|x|), so sinθ = |x| / sqrt(1 + x²)].

cosB = cos(π - θ) = -cosθ = -1 / sqrt(1 + x²).

Therefore, for general x:

sinB = |x| / sqrt(1 + x²), cosB = -sign(x) / sqrt(1 + x²). Wait, if x is positive, then B is in first quadrant. So previous analysis would be:

sinB = x / sqrt(1 + x²), cosB = 1 / sqrt(1 + x²). For x negative, sinB = |x| / sqrt(1 + x²) = -x / sqrt(1 + x²), cosB = -1 / sqrt(1 + x²).

Therefore, combining both cases:

sinB = x / sqrt(1 + x²) if x > 0, and sinB = -x / sqrt(1 + x²) if x < 0. But wait, for x < 0, |x| = -x, so sinB = |x| / sqrt(...) = -x / sqrt(...). So sinB = sign(x) * x / sqrt(1 + x²)? Wait, no. Wait:

For x positive: sinB = x / sqrt(1 + x²).

For x negative: sinB = (-x)/sqrt(1 + x²) = |x| / sqrt(...). Therefore, sinB = |x| / sqrt(1 + x²).

But if I write this as sinB = x / sqrt(1 + x²) if x > 0, and sinB = -x / sqrt(1 + x²) if x < 0. Therefore, sinB = (x / |x|) * |x| / sqrt(1 + x²) = x / sqrt(1 + x²) if x ≠ 0. Wait, no. Wait:

Wait, for x < 0: |x| = -x, so sinB = |x| / sqrt(1 + x²) = (-x)/sqrt(1 + x²). Which is the same as -x / sqrt(1 + x²). While for x positive, it's x / sqrt... Therefore, combining both, sinB = x / sqrt(1 + x²) when x > 0, and -x / sqrt(1 + x²) when x < 0. But since x / sqrt(1 + x²) is positive for x > 0, negative for x < 0, which matches.

But regardless, let me go back to the original expression:

sin(A + B) = sinA cosB + cosA sinB.

We have A = tan^{-1}(x), so sinA = x / sqrt(1 + x²), cosA = 1 / sqrt(1 + x²).

For B = cot^{-1}(1/x), we have sinB = x / sqrt(1 + x²) if x > 0, and sinB = -x / sqrt(1 + x²) if x < 0, and cosB = 1 / sqrt(1 + x²) if x > 0, and cosB = -1 / sqrt(1 + x²) if x < 0.

Wait, but for x < 0:

sinB = -x / sqrt(1 + x²), and cosB = -1 / sqrt(1 + x²). Therefore, sin(A + B) = sinA cosB + cosA sinB = [x / sqrt(...)] * [ -1 / sqrt(...) ] + [1 / sqrt(...)] * [ -x / sqrt(...) ] = [ -x / (1 + x²) ] + [ -x / (1 + x²) ] = -2x / (1 + x²).

Wait, but for x < 0, we have sin(A + B) = -2x / (1 + x²). Let me check this.

Wait, let me do it with the expressions. For x < 0:

sinA = x / sqrt(1 + x²) (which is negative)

cosB = -1 / sqrt(1 + x²)

Therefore, sinA * cosB = (x / sqrt(1 + x²)) * (-1 / sqrt(1 + x²)) = -x / (1 + x²)

cosA = 1 / sqrt(1 + x²)

sinB = -x / sqrt(1 + x²)

Therefore, cosA * sinB = (1 / sqrt(1 + x²)) * (-x / sqrt(1 + x²)) = -x / (1 + x²)

Therefore, total sin(A + B) = -x / (1 + x²) + (-x) / (1 + x²) = -2x / (1 + x²)

For x > 0, sinA = x / sqrt(...), cosB = 1 / sqrt(...), sinB = x / sqrt(...), cosA = 1 / sqrt(...)

Therefore, sinA * cosB + cosA * sinB = [x * 1 + 1 * x] / (1 + x²) = 2x / (1 + x²)

Therefore, for x > 0: sin(A + B) = 2x / (1 + x²)

For x < 0: sin(A + B) = -2x / (1 + x²)

Therefore, putting it all together:

sin(tan^{-1}(x) + cot^{-1}(1/x)) = { 2x / (1 + x²) if x > 0, -2x / (1 + x²) if x < 0 }

Therefore, the original equation:

sin(...) = 1/3 is equivalent to:

For x > 0: 2x / (1 + x²) = 1/3 => 6x = 1 + x² => x² - 6x + 1 = 0 => x = [6 ± sqrt(36 - 4)] / 2 = [6 ± sqrt(32)] / 2 = 3 ± 2√2. Both positive.

For x < 0: -2x / (1 + x²) = 1/3 => -2x = (1 + x²)/3 => -6x = 1 + x² => x² + 6x + 1 = 0 => x = [-6 ± sqrt(36 - 4)] / 2 = [-6 ± sqrt(32)] / 2 = [-6 ± 4√2]/2 = -3 ± 2√2.

But since x < 0, we check which solutions are negative:

-3 + 2√2 ≈ -3 + 2.828 ≈ -0.172 < 0.

-3 - 2√2 ≈ -5.828 < 0.

Therefore, for x < 0, solutions are x = -3 + 2√2 and x = -3 - 2√2.

Therefore, combining both cases, solutions are:

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

Which matches the previous result. Therefore, this confirms that there are four solutions.

Therefore, the answer is all four values. Therefore, boxed answers:

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

But the problem says "Enter all the solutions, separated by commas." So I need to present them as a list within a single box, separated by commas. Wait, the instructions say "put your final answer within \boxed{}". But usually, for multiple answers, we can write multiple boxed answers or separate them in one box. However, standard practice for multiple solutions is to list them all in one box separated by commas. For example:

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

But maybe as a single box with commas. Let me check previous examples. For example, if there are multiple solutions, they are often listed within a single \boxed{} separated by commas. For example, \boxed{-3 - 2\sqrt{2}}, \boxed{-3 + 2\sqrt{2}}, \boxed{3 - 2\sqrt{2}}, \boxed{3 + 2\sqrt{2}}. But instructions say "Enter all the solutions, separated by commas." So likely as a single box with commas.

Alternatively, sometimes problems expect solutions in a specific order or sorted. Let me check approximate values:

3 + 2√2 ≈ 5.828

3 - 2√2 ≈ 0.172

-3 + 2√2 ≈ -0.172

-3 - 2√2 ≈ -5.828

So sorted order: -5.828, -0.172, 0.172, 5.828. So solutions are -3 - 2√2, -3 + 2√2, 3 - 2√2, 3 + 2√2.

Therefore, final answer is these four values. So boxed as:

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

Alternatively, if the system allows multiple boxes, but likely the user wants them in a single box separated by commas. Since the original instruction says "put your final answer within \boxed{}", maybe they expect multiple boxed answers? However, standard practice on some platforms is to have a single box. Given the ambiguity, but following examples from other problems, sometimes multiple boxed answers are acceptable, but if only one box, then comma-separated inside.

Looking at the problem statement again: "Enter all the solutions, separated by commas." So likely they want comma-separated inside a single box. So:

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

But I need to check if this is acceptable. For example, in some contexts, they might expect ordered or simplified, but given that the problem didn't specify, it's safe to list them all.

Therefore, the final answer is the four solutions as above.

**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 if only one box is allowed:

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

But different platforms may vary. Since the user instruction says "put your final answer within \boxed{}", I think they expect multiple boxed answers. However, checking standard practice here, sometimes multiple boxed answers are used for multiple solutions. However, given the original instructions say "Enter all the solutions, separated by commas." and "put your final answer within \boxed{}", I think the expected format is all solutions in one box separated by commas. Therefore:

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

But to adhere to the instruction of one box, I think the correct way is:

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

However, I need to check if the system accepts multiple boxes or expects a single box. Since I can't be sure, but following the initial problem's examples, I think the correct way is multiple boxed answers. However, to conform to standard practice here, I'll put all in one box separated by commas.

**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 system might require only one box, I'll check again. In many online platforms, when multiple solutions are expected, they are put in one box separated by commas. For example, on AoPS or similar, the format is multiple boxed answers. However, given the original problem says "put your final answer within \boxed{}", I think they want one box with all solutions. Hence:

**Final Answer**
\boxed{-3 - 2\sqrt{2}, -3 + 2\sqrt{2}, 3 - 2\sqrt{2}, 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 argument of the sine function:
$$
\tan^{-1}(x) + \cot^{-1}\left( \frac{1}{x} \right).
$$

---

### Step 1: Simplify the Inverse Trigonometric Sum

We use the identity:
$$
\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}, \quad \text{for } x > 0.
$$

Also, recall that:
$$
\cot^{-1}\left( \frac{1}{x} \right) = \tan^{-1}(x), \quad \text{for } x > 0.
$$

Thus, when $ x > 0 $, the expression becomes:
$$
\tan^{-1}(x) + \tan^{-1}(x) = 2 \tan^{-1}(x).
$$

So we now rewrite the original equation for $ x > 0 $ as:
$$
\sin\left( 2 \tan^{-1}(x) \right) = \frac{1}{3}.
$$

Let $ \theta = \tan^{-1}(x) $, so $ \sin(2\theta) = \frac{1}{3} $.

---

### Step 2: Express in Terms of $ x $

Using the double angle identity for sine:
$$
\sin(2\theta) = 2 \sin(\theta) \cos(\theta).
$$

Given that $ \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}} $ and $ \cos(\tan^{-1}(x)) = \frac{1}{\sqrt{1 + x^2}} $, we compute:
$$
\sin(2 \tan^{-1}(x)) = \frac{2x}{1 + x^2}.
$$

Thus, the equation becomes:
$$
\frac{2x}{1 + x^2} = \frac{1}{3}.
$$

---

### Step 3: Solve the Resulting Quadratic Equation

Cross-multiplying:
$$
6x = 1 + x^2 \Rightarrow x^2 - 6x + 1 = 0.
$$

Solving:
$$
x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(1)}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}.
$$

So the two **positive** solutions are:
$$
x = 3 + 2\sqrt{2}, \quad x = 3 - 2\sqrt{2}.
$$

---

### Step 4: Consider the Negative Case

When $ x < 0 $, let $ x = -a $, where $ a > 0 $. Then:

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

Hence:
$$
\tan^{-1}(x) + \cot^{-1}(1/x) = -\tan^{-1}(a) + \pi - \tan^{-1}(a) = \pi - 2\tan^{-1}(a).
$$

Now use the identity $ \sin(\pi - \theta) = \sin(\theta) $, so:
$$
\sin(\pi - 2\tan^{-1}(a)) = \sin(2\tan^{-1}(a)) = \frac{1}{3}.
$$

This leads to the **same equation** as before:
$$
\frac{2a}{1 + a^2} = \frac{1}{3}.
$$

This results in the same quadratic:
$$
a^2 - 6a + 1 = 0 \Rightarrow a = 3 \pm 2\sqrt{2}.
$$

Since $ x = -a $, the **negative** solutions are:
$$
x = -3 - 2\sqrt{2}, \quad x = -3 + 2\sqrt{2}.
$$

---

### Step 5: Final Answer

All **four** real solutions are:

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

---

### Final Answer

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

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