Solve for `x: sin(2 tan^(-1)x)=1`

Answer

To solve the equation \( \sin(2 \tan^{-1} x) = 1 \), we can follow these steps:

### Step 1: Understand the equation
We know that \( \sin(y) = 1 \) when \( y = \frac{\pi}{2} + 2k\pi \) for any integer \( k \). In our case, we have:
\[ 2 \tan^{-1} x = \frac{\pi}{2} + 2k\pi \]

### Step 2: Solve for \( \tan^{-1} x \)
Dividing both sides of the equation by 2 gives:
\[ \tan^{-1} x = \frac{\pi}{4} + k\pi \]

### Step 3: Find \( x \)
Now, we can take the tangent of both sides to solve for \( x \):
\[ x = \tan\left(\frac{\pi}{4} + k\pi\right) \]

### Step 4: Simplify \( \tan\left(\frac{\pi}{4} + k\pi\right) \)
We know that \( \tan\left(\frac{\pi}{4}\right) = 1 \) and the tangent function has a period of \( \pi \). Therefore:
- For \( k = 0 \): \( x = \tan\left(\frac{\pi}{4}\right) = 1 \)
- For \( k = 1 \): \( x = \tan\left(\frac{5\pi}{4}\right) = 1 \) (since tangent is negative in the third quadrant)
- For \( k = -1 \): \( x = \tan\left(-\frac{3\pi}{4}\right) = 1 \) (since tangent is negative in the second quadrant)

Thus, the only solution for \( x \) is:
\[ x = 1 \]

### Final Answer
The solution for \( x \) is:
\[ \boxed{1} \]