Question

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. $\frac{1}{2}$
• D. $0$

Answer 1

The correct option is D $0$

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

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

We know that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

Replace $x$ by $1-x$

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

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

Let ${\sin }^{-1}x=a,$

Hence our equation becomes

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

$\Rightarrow \cos 2a=1-x$

$\Rightarrow 1-2{\sin }^{2}a=1-x$

$\Rightarrow 1-x=1-2{\sin }^{2}a$

$\Rightarrow 1-x=1-2x^2$

$\Rightarrow -2x^2+x=0$

$\Rightarrow -x(2x-1)=0$

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

But $x=\frac{1}{2}$ does not satisfy the equation

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

Put $x=\frac{1}{2}$ in L.H.S

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

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

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

$=\frac{-\pi }{6}$

$\ne \frac{\pi }{2}$

Hence $x=\frac{1}{2}$ not possible.

$\therefore x=0$ is the only solution.