The number of solution `(s)` of the equation `sin^(- 1)(1-x)-2sin^(- 1)x=pi/2` is/are

Options

A. 0
B. 1
C. 2
D. 3

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

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

### Step 3: Use the double angle formula
Using the double angle formula for cosine:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
Letting \( \theta = \sin^{-1}(x) \), we can write:
\[ \cos(2\sin^{-1}(x)) = 1 - 2x^2 \]
So, we substitute this into our equation:
\[ 1 - x = 1 - 2x^2 \]

### Step 4: Simplify the equation
Now, simplifying gives:
\[ -x = -2x^2 \]
This can be rearranged to:
\[ 2x^2 - x = 0 \]

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

### Step 6: Check the validity of solutions
Now we need to check if these solutions fall within the domain of the inverse sine function. The range of \( \sin^{-1}(x) \) is \( [-1, 1] \), so:
- For \( x = 0 \): \( \sin^{-1}(1-0) = \sin^{-1}(1) = \frac{\pi}{2} \) and \( 2\sin^{-1}(0) = 0 \). This satisfies the original equation.
- For \( x = \frac{1}{2} \): \( \sin^{-1}(1 - \frac{1}{2}) = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \) and \( 2\sin^{-1}(\frac{1}{2}) = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} \). Thus, \( \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \), which does not satisfy the equation.

### Conclusion
The only valid solution is \( x = 0 \). Therefore, the number of solutions \( s \) of the equation is:
\[ \boxed{1} \]