Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=\frac{\pi }{6}-2\times \frac{\pi }{6}=\frac{-\pi }{6}\ne \frac{\pi }{2}$

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

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