Solve `sin^(-1) (1 - x) - 2 sin^(-1) x = (pi)/(2)`

Options

A. `-1`
B. `0`
C. `1`
D. None of these

Explanation

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

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

### Step 2: Isolate \( \sin^{-1}(1 - x) \)
Rearranging the equation gives us:
\[ \sin^{-1}(1 - x) = \frac{\pi}{2} + 2 \sin^{-1}(x) \]

### Step 3: Apply the sine function
Taking the sine of both sides, we have:
\[ 1 - x = \sin\left(\frac{\pi}{2} + 2 \sin^{-1}(x)\right) \]

### Step 4: Use the sine addition formula
Using the sine addition formula, we know:
\[ \sin\left(\frac{\pi}{2} + \theta\right) = \cos(\theta) \]
Thus,
\[ 1 - x = \cos(2 \sin^{-1}(x)) \]

### Step 5: Simplify \( \cos(2 \sin^{-1}(x)) \)
Using the double angle formula for cosine, we have:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
Let \( \theta = \sin^{-1}(x) \), then:
\[ \cos(2 \sin^{-1}(x)) = 1 - 2x^2 \]
So, we can rewrite our equation as:
\[ 1 - x = 1 - 2x^2 \]

### Step 6: Rearrange the equation
Subtracting 1 from both sides gives:
\[ -x = -2x^2 \]
Multiplying through by -1 results in:
\[ x = 2x^2 \]

### Step 7: Rearranging to form a quadratic equation
Rearranging gives us:
\[ 2x^2 - x = 0 \]

### Step 8: Factor the equation
Factoring out \( x \):
\[ x(2x - 1) = 0 \]

### Step 9: Solve for \( x \)
Setting each factor to zero gives us:
1. \( x = 0 \)
2. \( 2x - 1 = 0 \) which simplifies to \( x = \frac{1}{2} \)

### Step 10: Verify the solutions
We need to check which of these solutions satisfy the original equation:
1. For \( x = 0 \):
\[ \sin^{-1}(1 - 0) - 2 \sin^{-1}(0) = \sin^{-1}(1) - 0 = \frac{\pi}{2} \]
This is valid.

2. For \( x = \frac{1}{2} \):
\[ \sin^{-1}(1 - \frac{1}{2}) - 2 \sin^{-1}(\frac{1}{2}) = \sin^{-1}(\frac{1}{2}) - 2 \cdot \frac{\pi}{6} = \frac{\pi}{6} - \frac{\pi}{3} = \frac{\pi}{6} - \frac{2\pi}{6} = -\frac{\pi}{6} \]
This does not satisfy the equation.

### Conclusion
The only solution is:
\[ \boxed{0} \]