Find the value of x which satisfy equation `2 tan^(-1) 2x = sin^(-1).(4x)/(1 + 4x^(2))`

Explanation

To solve the equation \( 2 \tan^{-1}(2x) = \sin^{-1}\left(\frac{4x}{1 + 4x^2}\right) \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2 \tan^{-1}(2x) = \sin^{-1}\left(\frac{4x}{1 + 4x^2}\right) \]

### Step 2: Use the double angle identity for sine
Recall the double angle identity for sine:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
We can express the right-hand side using the identity:
\[ \sin(2\theta) = \frac{4x}{1 + 4x^2} \]
where \( \theta = \tan^{-1}(2x) \).

### Step 3: Set \( 2x = \tan(\theta) \)
Let \( 2x = \tan(\theta) \). Then, we can express \( x \) as:
\[ x = \frac{1}{2} \tan(\theta) \]

### Step 4: Substitute into the equation
Substituting \( x \) back into the equation gives:
\[ 2 \tan^{-1}(2x) = 2\theta \]
Thus, we have:
\[ 2\theta = \sin^{-1}\left(\frac{4 \cdot \frac{1}{2} \tan(\theta)}{1 + 4\left(\frac{1}{2} \tan(\theta)\right)^2}\right) \]
This simplifies to:
\[ 2\theta = \sin^{-1}\left(\frac{2 \tan(\theta)}{1 + \tan^2(\theta)}\right) \]

### Step 5: Simplify the right-hand side
Using the identity \( \frac{\tan(\theta)}{1 + \tan^2(\theta)} = \sin(\theta) \):
\[ \sin^{-1}(\sin(2\theta)) = 2\theta \]
This holds true under the condition that \( 2\theta \) lies within the principal range of the sine function, which is \( -\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2} \).

### Step 6: Determine the range for \( \theta \)
From the condition \( -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \), we can find the corresponding range for \( x \):
\[ -\frac{\pi}{4} \leq \tan^{-1}(2x) \leq \frac{\pi}{4} \]

### Step 7: Convert back to \( x \)
Taking the tangent of all parts:
\[ \tan\left(-\frac{\pi}{4}\right) \leq 2x \leq \tan\left(\frac{\pi}{4}\right) \]
This gives:
\[ -1 \leq 2x \leq 1 \]
Dividing by 2:
\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \]

### Final Answer
Thus, the solution for \( x \) is:
\[ x \in \left[-\frac{1}{2}, \frac{1}{2}\right] \]