$\sin^{-1} (1 - x) - 2 \sin^{-1} x = \frac{\pi}{2}$, then x is equal to

(a) 0
(b) 1
(c) $\frac{1}{2}$
(d) 2

Choose the most appropriate answer from the options given below:

(a) only

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

Let $x = \sin \theta$

$\theta = \sin^{-1} x$

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

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

$1 - x = \cos 2 \theta$

$1 - \sin \theta = \cos 2 \theta$

⇒  $1 = \cos 2 \theta + \sin \theta$

$1 = 1 - 2 \sin^2 \theta + \sin \theta$

so  $2 \sin^2 \theta = \sin \theta$

$x = \sin \theta = 0, \frac{1}{2}$

at  $x = \frac{1}{2}$

LHS ≠ RHS

x = 0

Option: 3