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

(A) $0,\;\;\frac{1}{2}$

(B) $1,\;\;\frac{1}{2}$

(C) 0

(D) $\frac{1}{2}$

Option C is correct

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

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

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

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

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

But, $x = \frac{1}{2}$ does not satisfy the equation. So, x = 0.