If `sin^(-1)(2xsqrt(1-x^(2)))-2 sin^(-1) x=0` then x belongs to the interval

Options

A. `[-1,1]`
B. `[-1//sqrt(2),1sqrt(2)]`
C. `[-1,-1//sqrt(2)]`
D. `[1//sqrt(2),1]`

Answer

To solve the equation \( \sin^{-1}(2x\sqrt{1-x^2}) - 2\sin^{-1}(x) = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation:
\[ \sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}(x) \]

### Step 2: Letting \( x = \sin(\theta) \)
Let \( x = \sin(\theta) \). Then, we have:
\[ \sin^{-1}(x) = \theta \]

### Step 3: Substituting into the Equation
Substituting \( x = \sin(\theta) \) into the equation gives:
\[ \sin^{-1}(2\sin(\theta)\sqrt{1-\sin^2(\theta)}) = 2\theta \]
Since \( \sqrt{1-\sin^2(\theta)} = \cos(\theta) \), we can rewrite it as:
\[ \sin^{-1}(2\sin(\theta)\cos(\theta)) = 2\theta \]

### Step 4: Using the Double Angle Identity
Using the double angle identity, we know that:
\[ 2\sin(\theta)\cos(\theta) = \sin(2\theta) \]
Thus, we can rewrite our equation as:
\[ \sin^{-1}(\sin(2\theta)) = 2\theta \]

### Step 5: Analyzing the Inverse Function
The equation \( \sin^{-1}(\sin(2\theta)) = 2\theta \) holds true when \( 2\theta \) is in the range of \( \sin^{-1} \), which is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Therefore, we need:
\[ -\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2} \]
Dividing the entire inequality by 2 gives:
\[ -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \]

### Step 6: Relating Back to \( x \)
Since \( \theta = \sin^{-1}(x) \), we have:
\[ -\frac{\pi}{4} \leq \sin^{-1}(x) \leq \frac{\pi}{4} \]
Taking the sine of all parts of the inequality, we find:
\[ \sin\left(-\frac{\pi}{4}\right) \leq x \leq \sin\left(\frac{\pi}{4}\right) \]
This simplifies to:
\[ -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \]

### Conclusion
Thus, the solution to the equation is:
\[ x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \]