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 will follow these steps:

### Step 1: Rewrite the equation using the identity
We know that:
\[ \sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2} \]
Thus, we can rewrite \( \sin^{-1}(1-x) \) as:
\[ \sin^{-1}(1-x) = \frac{\pi}{2} - \cos^{-1}(1-x) \]
Substituting this into the original equation gives us:
\[ \frac{\pi}{2} - \cos^{-1}(1-x) - 2\sin^{-1}(x) = \frac{\pi}{2} \]
This simplifies to:
\[ -\cos^{-1}(1-x) - 2\sin^{-1}(x) = 0 \]
or
\[ \cos^{-1}(1-x) = -2\sin^{-1}(x) \]

### Step 2: Use the cosine identity
We know that \( \cos(-\theta) = \cos(\theta) \), so:
\[ \cos^{-1}(1-x) = 2\sin^{-1}(x) \]
Taking the cosine of both sides gives us:
\[ 1-x = \cos(2\sin^{-1}(x)) \]

### Step 3: Apply the double angle formula
Using the double angle formula for cosine:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
Let \( \theta = \sin^{-1}(x) \). Then:
\[ \cos(2\sin^{-1}(x)) = 1 - 2\sin^2(\sin^{-1}(x)) = 1 - 2x^2 \]
So we have:
\[ 1 - x = 1 - 2x^2 \]

### Step 4: Rearranging the equation
Now, rearranging gives:
\[ -x = -2x^2 \]
or
\[ 2x^2 - x = 0 \]

### Step 5: Factor the equation
Factoring out \( x \):
\[ x(2x - 1) = 0 \]
This gives us two solutions:
\[ x = 0 \quad \text{or} \quad 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \]

### Step 6: Check the solutions
1. For \( x = 0 \):
\[ \sin^{-1}(1-0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 0 = \frac{\pi}{2} \]
This satisfies the equation.

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:
\[ \boxed{0} \]