Solve for $x$: $\sin^{-1}(1 - x) - 2 \sin^{-1} x = \frac{\pi}{2}$.

$0$

The correct answer is Option (3) → 0 ##

$\sin^{-1}(1 - x) - 2 \sin^{-1} x = \frac{\pi}{2}$

$\Rightarrow (1 - x) = \sin \left( \frac{\pi}{2} + 2 \sin^{-1} x \right)$

$\Rightarrow (1 - x) = \cos(2 \sin^{-1} x)$

$\Rightarrow 1 - x = 1 - 2x^2$

$\Rightarrow 2x^2 - x = 0$

$\Rightarrow x = 0, x = \frac{1}{2}$

Since $x = \frac{1}{2}$ does not satisfy the given equation,

$∴x = 0$ is the required solution.