Question

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

• A. $-\frac{1}{2}$
• B. $1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$0$

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 $2{\sin }^{-1}x=t$

$\therefore x=\sin \frac{t}{2}=\sqrt{\frac{1-\cos t}{2}}$

$\Rightarrow x^2=\frac{1-\cos t}{2}$
$\Rightarrow \cos t=1-2x^2$

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

Substitute this value in the obtained expression,
$\Rightarrow 1-x=\cos \left[{\cos }^{-1}\right(1-2x^2\left)\right]$
$\Rightarrow 1-x=1-2x^2$
$\Rightarrow 2x^2-x=0$
$\Rightarrow x(2x-1)=0$
$\therefore x=0$ or $x=\frac{1}{2}$
For $x=\frac{1}{2}$,
${\sin }^{-1}(1-x)-2{\sin }^{-1}x={\sin }^{-1}\left(\frac{1}{2}\right)-2{\sin }^{-1}\left(\frac{1}{2}\right)$

$=-{\sin }^{-1}\left(\frac{1}{2}\right)$

$=-\frac{\pi }{6}$
So, $x=\frac{1}{2}$ is not the solution of the given equation.
For $x=0$,
${\sin }^{-1}(1-x)-2{\sin }^{-1}x={\sin }^{-1}\left(1\right)-2{\sin }^{-1}\left(0\right)$

$=\frac{\pi }{2}-0$

$=\frac{\pi }{2}$
Hence, the correct answer from the given alternatives is option $C$.