The equation `2 cos^(-1)x=sin^(-1)(2x sqrt(1-x^(2)))` is valid for all values of x satisfying

Options

A. `-1 le x le 1`
B. `0 le x le 1`
C. `0 le x le 1// sqrt(2)`
D. `1//sqrt(2) le x le 1`

Answer

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

### Step 1: Let \( \theta = \cos^{-1} x \)
From this, we can express \( x \) in terms of \( \theta \):
\[ x = \cos \theta \]

### Step 2: Substitute \( x \) in the equation
Substituting \( x = \cos \theta \) into the equation gives:
\[ 2\theta = \sin^{-1}(2 \cos \theta \sqrt{1 - \cos^2 \theta}) \]

### Step 3: Simplify the right-hand side
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:
\[ \sqrt{1 - \cos^2 \theta} = \sin \theta \]
Thus, the right-hand side becomes:
\[ \sin^{-1}(2 \cos \theta \sin \theta) \]

### Step 4: Use the double angle identity
Recall that \( 2 \cos \theta \sin \theta = \sin(2\theta) \). Therefore, we can rewrite the equation as:
\[ 2\theta = \sin^{-1}(\sin(2\theta)) \]

### Step 5: Analyze the sine inverse function
The function \( \sin^{-1}(y) \) gives us \( y \) in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\). Therefore, for \( \sin^{-1}(\sin(2\theta)) \) to equal \( 2\theta \), we need:
\[ 2\theta = 2n\pi + (-1)^n(2\theta) \quad \text{for some integer } n \]
This means \( 2\theta \) must lie within the range of \( \sin^{-1} \).

### Step 6: Determine the range of \( \theta \)
Since \( \theta = \cos^{-1} x \), it follows that:
\[ 0 \leq \theta \leq \pi \]
Thus, \( 0 \leq 2\theta \leq 2\pi \).

### Step 7: Find the valid range for \( x \)
For \( 2\theta \) to remain within the range of \( \sin^{-1} \), we need:
\[ -1 \leq 2 \cos \theta \sqrt{1 - \cos^2 \theta} \leq 1 \]
This condition is satisfied when \( x \) lies within the interval:
\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \]

### Conclusion
The equation \( 2 \cos^{-1} x = \sin^{-1}(2x \sqrt{1 - x^2}) \) is valid for all values of \( x \) satisfying:
\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \]