If `sin^(-1)(1-x)-2 sin^(-1)x=(pi)/(2)`, then `x` is equal to

Options

A. `0,(1)/(2)`
B. `1,(1)/(2)`
C. `0`
D. `(1)/(2)`

Explanation

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

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate \( \sin^{-1}(1-x) \):
\[ \sin^{-1}(1-x) = \frac{\pi}{2} + 2 \sin^{-1}(x) \]

**Hint:** When rearranging equations, always aim to isolate the term involving the inverse function.

### Step 2: Applying the Sine Function
Next, we apply the sine function to both sides of the equation:
\[ 1 - x = \sin\left(\frac{\pi}{2} + 2 \sin^{-1}(x)\right) \]
Using the identity \( \sin\left(\frac{\pi}{2} + \theta\right) = \cos(\theta) \), we have:
\[ 1 - x = \cos(2 \sin^{-1}(x)) \]

**Hint:** Remember the sine and cosine identities when dealing with angles.

### Step 3: Using the Double Angle Formula
Now, we can use the double angle formula for cosine:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
Let \( \theta = \sin^{-1}(x) \), then:
\[ \cos(2 \sin^{-1}(x)) = 1 - 2x^2 \]
Thus, we can rewrite our equation as:
\[ 1 - x = 1 - 2x^2 \]

**Hint:** Familiarize yourself with trigonometric identities to simplify expressions.

### Step 4: Simplifying the Equation
Now, we simplify the equation:
\[ 1 - x = 1 - 2x^2 \]
Subtracting 1 from both sides:
\[ -x = -2x^2 \]
Multiplying through by -1 gives:
\[ x = 2x^2 \]

**Hint:** When simplifying equations, ensure to keep the signs consistent.

### Step 5: Rearranging to Form a Quadratic Equation
Rearranging gives us:
\[ 2x^2 - x = 0 \]
Factoring out \( x \):
\[ x(2x - 1) = 0 \]

**Hint:** Factoring is a useful technique for solving polynomial equations.

### Step 6: Solving for \( x \)
Setting each factor to zero gives us:
1. \( x = 0 \)
2. \( 2x - 1 = 0 \) which leads to \( x = \frac{1}{2} \)

**Hint:** Always check the solutions against the original equation.

### Step 7: Checking the Solutions
Now we need to check both solutions in 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 \( x = \frac{1}{2} \):
\[ \sin^{-1}(1 - \frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \sin^{-1}(\frac{1}{2}) - 2 \cdot \frac{\pi}{6} = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \]
This does not equal \( \frac{\pi}{2} \).

Thus, the only valid solution is:
\[ \boxed{0} \]