Solve for x :
`3 sin^(-1) ((2x)/(1+x^(2))) - 4 cos^(-1) ((1-x^(2))/(1+x^(2))) +2 tan^(-1). ((2x)/(1-x^(2))) = pi/ 2 . `

Answer

To solve the equation

\[ 3 \sin^{-1} \left( \frac{2x}{1+x^2} \right) - 4 \cos^{-1} \left( \frac{1-x^2}{1+x^2} \right) + 2 \tan^{-1} \left( \frac{2x}{1-x^2} \right) = \frac{\pi}{2}, \]

we can start by using the identities for the inverse trigonometric functions.

### Step 1: Simplifying the Inverse Functions

1. **Recognize the identities:**
- The expression \(\sin^{-1} \left( \frac{2x}{1+x^2} \right)\) can be rewritten using the double angle formula for tangent:
\[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) = 2 \tan^{-1}(x). \]
- The expression \(\cos^{-1} \left( \frac{1-x^2}{1+x^2} \right)\) can also be rewritten:
\[ \cos^{-1} \left( \frac{1-x^2}{1+x^2} \right) = 2 \tan^{-1}(x). \]
- The expression \(\tan^{-1} \left( \frac{2x}{1-x^2} \right)\) is already in a suitable form:
\[ \tan^{-1} \left( \frac{2x}{1-x^2} \right) = 2 \tan^{-1}(x). \]

### Step 2: Substitute the Identities

Now substituting these identities into the original equation:

\[ 3(2 \tan^{-1}(x)) - 4(2 \tan^{-1}(x)) + 2(2 \tan^{-1}(x)) = \frac{\pi}{2}. \]

This simplifies to:

\[ 6 \tan^{-1}(x) - 8 \tan^{-1}(x) + 4 \tan^{-1}(x) = \frac{\pi}{2}. \]

### Step 3: Combine Like Terms

Combining the terms gives:

\[ (6 - 8 + 4) \tan^{-1}(x) = \frac{\pi}{2}. \]

This simplifies to:

\[ 2 \tan^{-1}(x) = \frac{\pi}{2}. \]

### Step 4: Solve for \(\tan^{-1}(x)\)

Dividing both sides by 2:

\[ \tan^{-1}(x) = \frac{\pi}{4}. \]

### Step 5: Find \(x\)

Taking the tangent of both sides, we find:

\[ x = \tan\left(\frac{\pi}{4}\right) = 1. \]

### Conclusion

Thus, the solution to the equation is:

\[ \boxed{1}. \]