If `sin ^(-1) (1-x)-2 sin ^(-1) x = (pi)/(2)` then x = ?

Options

A. `0,(1)/(2),1`
B. `0,(1)/(2)`
C. 0
D. None of these

Answer

To solve the equation \( \sin^{-1}(1-x) - 2 \sin^{-1}(x) = \frac{\pi}{2} \), we can 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 the Inverse Sine Function
Rearranging the equation gives:
\[ \sin^{-1}(1-x) = \frac{\pi}{2} + 2 \sin^{-1}(x) \]

### Step 3: Apply the Sine Function
Now, we apply the sine function to both sides:
\[ 1 - x = \sin\left(\frac{\pi}{2} + 2 \sin^{-1}(x)\right) \]

### Step 4: Use the Sine Addition Formula
Using the identity \( \sin\left(\frac{\pi}{2} + \theta\right) = \cos(\theta) \):
\[ 1 - x = \cos(2 \sin^{-1}(x)) \]

### Step 5: Use the Cosine Double Angle Identity
We know that \( \cos(2\theta) = 1 - 2\sin^2(\theta) \). Thus:
\[ 1 - x = 1 - 2\sin^2(\sin^{-1}(x)) \]
Since \( \sin(\sin^{-1}(x)) = x \):
\[ 1 - x = 1 - 2x^2 \]

### Step 6: Simplify the Equation
Now, we can simplify:
\[ -x = -2x^2 \]
This leads to:
\[ 2x^2 - x = 0 \]

### Step 7: Factor the Equation
Factoring out \( x \):
\[ x(2x - 1) = 0 \]

### Step 8: 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 9: Verify Solutions
We need to check which of these solutions satisfy the original equation:
- For \( x = 0 \):
\[ \sin^{-1}(1-0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 0 = \frac{\pi}{2} \quad \text{(valid)} \]
- For \( x = \frac{1}{2} \):
\[ \sin^{-1}(1 - \frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \sin^{-1}(\frac{1}{2}) - 2 \cdot \frac{\pi}{6} = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \quad \text{(not valid)} \]

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