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

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

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

### Step 3: Apply the Sine Function
Taking the sine of both sides, we have:
\[ 1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1}(x)\right) \]
Using the sine addition formula, we know that:
\[ \sin\left(\frac{\pi}{2} + \theta\right) = \cos(\theta) \]
Thus, we can rewrite the equation as:
\[ 1 - x = \cos(2\sin^{-1}(x)) \]

### Step 4: Use the Cosine Double Angle Identity
Using the identity \( \cos(2\theta) = 1 - 2\sin^2(\theta) \), we substitute \( \theta = \sin^{-1}(x) \):
\[ \cos(2\sin^{-1}(x)) = 1 - 2x^2 \]
So, we have:
\[ 1 - x = 1 - 2x^2 \]

### Step 5: Simplify the Equation
Cancelling \( 1 \) from both sides gives:
\[ -x = -2x^2 \]
Rearranging this leads to:
\[ 2x^2 - x = 0 \]

### Step 6: Factor the Quadratic Equation
Factoring out \( x \):
\[ x(2x - 1) = 0 \]

### Step 7: Solve for \( x \)
Setting each factor to zero gives:
1. \( x = 0 \)
2. \( 2x - 1 = 0 \) which gives \( x = \frac{1}{2} \)

### Step 8: Check the Solutions
We need to check which of these solutions satisfies the original equation:
- For \( x = 0 \):
\[ \sin^{-1}(1-0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 0 = \frac{\pi}{2} \]
This satisfies the equation.

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

### Final Solution
Thus, the only solution is:
\[ \boxed{0} \]