Question

Find values of x which satisfy : ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$.

• A. $x=0$
• B. $x=1/2$
• C. $,x=0andx=1/2$
• D. None of these

Answer 1

The correct option is A $x=0$

Given, ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\pi /2$

Let $x=\sin y$

Therefore, the given equation reduces to
${\sin }^{-1}(1-\sin y)-2{\sin }^{-1}\sin y=\frac{\pi }{2}$

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

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

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

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

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

$2{\sin }^{2}y=\sin y$

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

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

$\Rightarrow$ $\sin y=0$ or $1/2$

$\Rightarrow x=0$ or $1/2$

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

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

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

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

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

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

Thus, $x=0$ is only solution.