If `tan^(-1)x+2cot^(-1)x=(5pi)/6`, then find x.

Answer

To solve the equation \( \tan^{-1}x + 2\cot^{-1}x = \frac{5\pi}{6} \), we will follow these steps:

### Step 1: Rewrite the Equation
We know that:
\[ \cot^{-1}x = \frac{\pi}{2} - \tan^{-1}x \]
Substituting this into the equation gives:
\[ \tan^{-1}x + 2\left(\frac{\pi}{2} - \tan^{-1}x\right) = \frac{5\pi}{6} \]

### Step 2: Simplify the Equation
Now, simplify the left-hand side:
\[ \tan^{-1}x + \pi - 2\tan^{-1}x = \frac{5\pi}{6} \]
This simplifies to:
\[ \pi - \tan^{-1}x = \frac{5\pi}{6} \]

### Step 3: Isolate \( \tan^{-1}x \)
Rearranging gives:
\[ -\tan^{-1}x = \frac{5\pi}{6} - \pi \]
This simplifies to:
\[ -\tan^{-1}x = \frac{5\pi}{6} - \frac{6\pi}{6} = -\frac{\pi}{6} \]
Thus, we have:
\[ \tan^{-1}x = \frac{\pi}{6} \]

### Step 4: Solve for \( x \)
Taking the tangent of both sides:
\[ x = \tan\left(\frac{\pi}{6}\right) \]
We know that:
\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Thus, we find:
\[ x = \frac{1}{\sqrt{3}} \]

### Final Answer:
The value of \( x \) is:
\[ \boxed{\frac{1}{\sqrt{3}}} \]