Okay, so I need to solve the equation sin(arctan(x) + arccot(1/x)) = 1/3. Hmm, let's see. First, let me recall some trigonometric identities and inverse trigonometric function properties. Maybe I can simplify the argument of the sine function first.

Let me denote arctan(x) as θ. So, θ = arctan(x), which means tanθ = x. Similarly, arccot(1/x) can be written as φ, where φ = arccot(1/x), so cotφ = 1/x. But cotφ is the reciprocal of tanφ, so tanφ = x. Wait a second, that's interesting. So both θ and φ have tangent equal to x? But θ is arctan(x) and φ is arccot(1/x). Let me check if that's correct.

If φ = arccot(1/x), then cotφ = 1/x, which means tanφ = x. So yes, both θ and φ have tan equal to x. But arctan(x) and arccot(1/x) are related. Let me recall that arccot(z) = arctan(1/z) if z > 0. Wait, so arccot(1/x) would be arctan(x) if x > 0. But is that the case? Let me verify.

Actually, arccot(z) is defined as the angle whose cotangent is z. Since cotangent is the reciprocal of tangent, arccot(z) = arctan(1/z) when z > 0. So, arccot(1/x) = arctan(x) when 1/x > 0, which is when x > 0. If x < 0, then arccot(1/x) would be arctan(x) shifted by π, right? Because arccot(z) is in the range (0, π), whereas arctan(x) is in (-π/2, π/2). So, if x is negative, arccot(1/x) would be π + arctan(x). Wait, is that correct?

Let me think. Suppose x is negative. Then 1/x is negative. So arccot(1/x) would be the angle in (0, π) whose cotangent is 1/x (negative). Since cotangent is negative in the second quadrant, where angles are between π/2 and π. So arccot(1/x) for x negative would be π + arctan(1/x). But arctan(1/x) is negative because 1/x is negative. So, arccot(1/x) = π + arctan(1/x) = π - |arctan(1/x)|. Alternatively, arccot(z) for z negative is π + arctan(z). Wait, maybe I need to recall the exact relationship.

Wait, arccot(z) is defined as the angle φ in (0, π) such that cotφ = z. So if z is negative, then φ is in (π/2, π). So, for z negative, arccot(z) = π + arctan(z). Because arctan(z) for z negative is in (-π/2, 0). So π + arctan(z) would be in (π/2, π). So yes, arccot(z) = π + arctan(z) when z < 0. Therefore, arccot(1/x) = π + arctan(1/x) if 1/x < 0, which is when x < 0. If x > 0, then arccot(1/x) = arctan(x). If x = 0, then arccot(1/x) is undefined since 1/x would be infinity.

So, putting this together:

If x > 0, then arccot(1/x) = arctan(x). Therefore, the argument of the sine function becomes arctan(x) + arctan(x) = 2 arctan(x). So sin(2 arctan(x)) = 1/3.

If x < 0, then arccot(1/x) = π + arctan(1/x). Since 1/x is negative here, arctan(1/x) is negative. Let's denote arctan(1/x) as -α, where α > 0. Then arccot(1/x) = π - α. Therefore, the argument becomes arctan(x) + arccot(1/x) = arctan(x) + π - α. But arctan(x) is negative since x < 0, let's say arctan(x) = -β, where β > 0. Then the argument is -β + π - α. But α = arctan(1/x) = arctan(-1/|x|) = -arctan(1/|x|) = -α', so α = -α' where α' = arctan(1/|x|). Therefore, the argument is -β + π - (-α') = π - β + α'. But β = arctan(x) = arctan(-|x|) = -arctan(|x|) = -β', so β = -β', so -β = β'. Then the argument becomes π + β' + α' = π + (arctan(|x|) + arctan(1/|x|)).

Wait, this seems a bit convoluted. Maybe there's a better way to approach this.

Alternatively, since arccot(1/x) = arctan(x) when x > 0, and arccot(1/x) = π + arctan(x) when x < 0. Let me verify this with an example. Suppose x = 1. Then arccot(1/1) = arccot(1) = π/4, and arctan(1) = π/4. So yes, arccot(1/x) = arctan(x) when x > 0. If x = -1, then arccot(1/(-1)) = arccot(-1). Cotangent of π/4 is 1, but cotangent of 3π/4 is -1. So arccot(-1) = 3π/4. arctan(-1) = -π/4. So 3π/4 = π + (-π/4) = 3π/4. So yes, arccot(1/x) = π + arctan(x) when x < 0.

Therefore, the argument inside the sine function is:

If x > 0: 2 arctan(x)

If x < 0: arctan(x) + π + arctan(x) = π + 2 arctan(x)

So, the equation becomes:

If x > 0: sin(2 arctan(x)) = 1/3

If x < 0: sin(π + 2 arctan(x)) = 1/3

But sin(π + θ) = -sinθ, so sin(π + 2 arctan(x)) = -sin(2 arctan(x)). Therefore, for x < 0, we have -sin(2 arctan(x)) = 1/3 => sin(2 arctan(x)) = -1/3.

So, summarizing:

For x > 0: sin(2 arctan(x)) = 1/3

For x < 0: sin(2 arctan(x)) = -1/3

Therefore, regardless of the sign of x, we can solve sin(2 arctan(x)) = ±1/3. Let's handle both cases.

First, let's let θ = arctan(x). Then tanθ = x. So 2θ is the argument of the sine function. Therefore, sin(2θ) = 2 sinθ cosθ. Since tanθ = x, we can express sinθ and cosθ in terms of x.

Let me recall that if tanθ = x, then we can imagine a right triangle where the opposite side is x and the adjacent side is 1, so the hypotenuse is sqrt(1 + x²). Therefore, sinθ = x / sqrt(1 + x²) and cosθ = 1 / sqrt(1 + x²). Therefore, sin(2θ) = 2*(x / sqrt(1 + x²))*(1 / sqrt(1 + x²)) = 2x / (1 + x²).

Therefore, sin(2 arctan(x)) = 2x / (1 + x²). Therefore, our equation becomes:

For x > 0: 2x / (1 + x²) = 1/3

For x < 0: 2x / (1 + x²) = -1/3

So, we can solve these two equations separately.

First, let's solve 2x / (1 + x²) = 1/3 for x > 0.

Multiply both sides by (1 + x²):

2x = (1 + x²)/3

Multiply both sides by 3:

6x = 1 + x²

Bring all terms to one side:

x² - 6x + 1 = 0

This is a quadratic equation. Using the quadratic formula:

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

But since we're solving for x > 0, both solutions 3 + 2√2 and 3 - 2√2 need to be checked. Let's compute 3 - 2√2. Since √2 ≈ 1.414, 2√2 ≈ 2.828. So 3 - 2.828 ≈ 0.172, which is positive. So both roots are positive. Therefore, x = 3 + 2√2 and x = 3 - 2√2.

Now, check if these satisfy the original equation. Let's plug x = 3 + 2√2 into 2x / (1 + x²):

First, compute x²:

(3 + 2√2)² = 9 + 12√2 + 8 = 17 + 12√2

Then 2x = 2*(3 + 2√2) = 6 + 4√2

So 2x / (1 + x²) = (6 + 4√2)/(1 + 17 + 12√2) = (6 + 4√2)/(18 + 12√2)

Factor numerator and denominator:

Numerator: 2*(3 + 2√2)

Denominator: 6*(3 + 2√2)

Therefore, (6 + 4√2)/(18 + 12√2) = (2*(3 + 2√2))/(6*(3 + 2√2)) = 2/6 = 1/3. Correct.

Similarly, x = 3 - 2√2. Compute x²:

(3 - 2√2)² = 9 - 12√2 + 8 = 17 - 12√2

2x = 2*(3 - 2√2) = 6 - 4√2

Denominator: 1 + x² = 1 + 17 - 12√2 = 18 - 12√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, (6 - 4√2)/(18 - 12√2) = (2*(3 - 2√2))/(6*(3 - 2√2)) = 2/6 = 1/3. Correct.

So both solutions are valid for x > 0.

Now, for x < 0: 2x / (1 + x²) = -1/3

Multiply both sides by (1 + x²):

2x = (-1/3)(1 + x²)

Multiply both sides by 3:

6x = -1 - x²

Bring all terms to one side:

x² + 6x + 1 = 0

Quadratic equation: x = [-6 ± sqrt(36 - 4*1*1)] / 2 = [-6 ± sqrt(32)] / 2 = [-6 ± 4√2]/2 = -3 ± 2√2

Since x < 0, we need to check which of these solutions are negative.

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

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

Therefore, both roots are negative. Let's check if they satisfy the original equation.

First, x = -3 + 2√2 ≈ -0.172. Compute 2x / (1 + x²):

2x = 2*(-3 + 2√2) = -6 + 4√2 ≈ -6 + 5.656 ≈ -0.344

x² = (-3 + 2√2)² = 9 - 12√2 + 8 = 17 - 12√2 ≈ 17 - 16.97 ≈ 0.03

1 + x² ≈ 1 + 0.03 ≈ 1.03

So 2x / (1 + x²) ≈ -0.344 / 1.03 ≈ -0.334, which is approximately -1/3 (since -1/3 ≈ -0.333...). So this is correct.

Similarly, x = -3 - 2√2 ≈ -5.828. Compute 2x / (1 + x²):

2x = 2*(-3 - 2√2) = -6 - 4√2 ≈ -6 - 5.656 ≈ -11.656

x² = (-3 - 2√2)² = 9 + 12√2 + 8 = 17 + 12√2 ≈ 17 + 16.97 ≈ 33.97

1 + x² ≈ 34.97

So 2x / (1 + x²) ≈ -11.656 / 34.97 ≈ -0.333 ≈ -1/3. Correct.

Therefore, both solutions are valid for x < 0.

Therefore, all solutions are x = 3 ± 2√2 and x = -3 ± 2√2. Wait, but hold on. Wait, the quadratic for x > 0 gave x = 3 ± 2√2, which are both positive. For x < 0, the solutions are -3 ± 2√2. But -3 + 2√2 is approximately -0.172, which is negative, and -3 - 2√2 is approximately -5.828, which is also negative. Therefore, all four solutions are distinct.

But let me check if all four solutions are valid. Sometimes when solving equations, especially trigonometric ones, extraneous solutions can appear, but in this case, since we converted the equation into algebraic equations and the steps were reversible, all solutions should be valid. Let me verify with the original equation.

Take x = 3 + 2√2:

Compute arctan(x) + arccot(1/x). Since x > 0, arccot(1/x) = arctan(x). So the argument is 2 arctan(x). Then sin(2 arctan(x)) = 2x/(1 + x²) = 1/3, which matches. So correct.

Similarly, x = 3 - 2√2:

Again, x > 0, so arccot(1/x) = arctan(x). The argument is 2 arctan(x), sin(2 arctan(x)) = 1/3. Correct.

For x = -3 + 2√2 ≈ -0.172:

Compute arctan(x) + arccot(1/x). Since x < 0, arccot(1/x) = π + arctan(1/x). As calculated earlier, this becomes π + arctan(1/x) + arctan(x). But arctan(x) is negative, let's say arctan(x) = -α, so arccot(1/x) = π - α. Therefore, the argument is -α + π - α = π - 2α. Then sin(π - 2α) = sin(2α). But sin(π - θ) = sinθ, so sin(π - 2α) = sin(2α). Wait, but in the original equation, sin(arctan(x) + arccot(1/x)) = 1/3. For x < 0, this is sin(π - 2α) = 1/3. So sin(π - 2α) = sin(2α) = 1/3. But according to our previous derivation, sin(2 arctan(x)) = sin(-2α) = -sin(2α). Wait, maybe I need to check again.

Wait, if x < 0, then arctan(x) = -α, where α = |arctan(x)|. Then arccot(1/x) = π + arctan(1/x). But 1/x is negative, so arctan(1/x) = -β, where β = |arctan(1/x)|. Therefore, arccot(1/x) = π - β. Therefore, the argument is -α + π - β. But since α and β are related by x and 1/x. Let's see, x = tan(-α) = -tan(α), so tan(α) = -x. Similarly, 1/x = -cot(α) = -tan(π/2 - α). Wait, maybe this is getting too convoluted.

Alternatively, since we already converted the equation into 2x/(1 + x²) = -1/3 for x < 0, and found solutions x = -3 ± 2√2, which satisfy 2x/(1 + x²) = -1/3, and since the original equation reduces to this, all these solutions are valid. Therefore, all four solutions are valid.

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

But let me check if x = -3 + 2√2 is positive or negative. Since √2 ≈ 1.414, 2√2 ≈ 2.828. Therefore, -3 + 2.828 ≈ -0.172, which is negative. So x = -3 + 2√2 is negative, and x = -3 - 2√2 is also negative, as -3 - 2√2 ≈ -5.828. So the four solutions are two positive and two negative. However, in the initial split, we considered x > 0 and x < 0, so all four solutions are valid.

Therefore, the final solutions are x = 3 + 2√2, x = 3 - 2√2, x = -3 + 2√2, and x = -3 - 2√2. Let me write them in order:

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

Alternatively, combining the positive and negative solutions:

Positive solutions: 3 ± 2√2

Negative solutions: -3 ± 2√2

So, all four solutions are distinct and valid. Therefore, the answer is x = 3 + 2√2, 3 - 2√2, -3 + 2√2, -3 - 2√2.

**Final Answer**
The solutions are \(\boxed{3 + 2\sqrt{2}}\), \(\boxed{3 - 2\sqrt{2}}\), \(\boxed{-3 + 2\sqrt{2}}\), and \(\boxed{-3 - 2\sqrt{2}}\).
Thus, the solutions are \( x = \boxed{3 + 2\sqrt{2}} \), \( x = \boxed{3 - 2\sqrt{2}} \), \( x = \boxed{-3 + 2\sqrt{2}} \), and \( x = \boxed{-3 - 2\sqrt{2}} \).
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