Question

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

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

Answer 1

Answer: (C)

$0$

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

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

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

Let ${\sin }^{-1}x=\theta \Rightarrow \sin \theta =x\Rightarrow \cos \theta =\sqrt{1-x^2}$

$\therefore \theta ={\cos }^{-1}\left(\sqrt{1-x^2}\right)$

$\therefore {\sin }^{-1}x={\cos }^{-1}\left(\sqrt{1-x^2}\right)$

Therefore, from equation (1), we have
$-2{\cos }^{-1}\left(\sqrt{1-x^2}\right)={\cos }^{-1}(1-x)$

Put $x=\sin y$

$-2{\cos }^{-1}\left(\sqrt{1-{\sin }^{2}y}\right)={\cos }^{-1}(1-\sin y)$

$\Rightarrow -2{\cos }^{-1}\left(\cos y\right)={\cos }^{-1}(1-\sin y)$

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

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

$\Rightarrow 2{\sin }^{2}y-\sin y=0$

$\Rightarrow \sin y(2\sin y-1)=0$

$\Rightarrow \sin y=0or\frac{1}{2}$

$\therefore x=0orx=\frac{1}{2}$

But, when $x=\frac{1}{2}$, it can be observed that :

$\text{L.H.S.}={sin}^{-1}(1-\frac{1}{2})-2{\sin }^{-1}\frac{1}{2}$

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

$=-{\sin }^{-1}\frac{1}{2}$

$=-\frac{\pi }{6}\ne \frac{\pi }{2}\ne \text{R.H.S.}$

$\therefore x=\frac{1}{2}$ is not the solution of the given equation.

Thus , $x=0.$

Hence, the correct answer is C.