Explanation

(c)
Given, $\sin^{- 1}{(1 - x) = \frac{\pi}{2} + 2\sin^{- 1}x}$
$\Rightarrow \ \ \ 1 - x = \sin\left( \frac{\pi}{2} + 2\sin^{- 1}x \right)$
$\Rightarrow \ \ \ 1 - x = \cos{(2\sin^{- 1}x)}$
$\Rightarrow \ \ 1 - x = \cos{(2\cos^{- 1}\sqrt{1 - x^{2}})}$
$\Rightarrow \ \ 1 - x = \cos{\{\cos^{- 1}{(1 - 2x^{2})}\}}$
$\Rightarrow \ \ 1 - x = 1 - 2x^{2}$
$\Rightarrow \ \ x = 0,\frac{1}{2}$
$\Rightarrow \ \ \ x = 0\ \ \left\lbrack \because x = \frac{1}{2}\ does\ not\ satisfy\ the\ given\ equation \right\rbrack$