Question

The number of value(s) of $x$ satisfying ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$ is

• A. $0$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

We have,
${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}(1-x)=\frac{\pi }{2}+2{\sin }^{-1}x$
$\Rightarrow 1-x=\sin (\frac{\pi }{2}+2{\sin }^{-1}x)$
$\Rightarrow 1-x=\cos \left(2{\sin }^{-1}x\right)$
(Let ${\sin }^{-1}x=\theta \Rightarrow x=\sin \theta$)
$\Rightarrow 1-\sin \theta =\cos 2\theta =1-2{\sin }^2\theta$
$\Rightarrow x=\sin \theta =0,\frac{1}{2}$

For $x=\frac{1}{2},$
${\sin }^{-1}(1-x)-2{\sin }^{-1}x$
$={\sin }^{-1}\frac{1}{2}-2{\sin }^{-1}\frac{1}{2}=-{\sin }^{-1}\frac{1}{2}=-\frac{\pi }{6}\ne \frac{\pi }{2}$
So, $x=\frac{1}{2}$ is not a root of the given equation.

Clearly, $x=0$ satisfies the equation.
Hence, $x=0$ is a root of the given equation.