Question

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

Answer 1

${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$ ....(1)
${\sin }^{-1}(1-x)=\frac{\pi }{2}+2{\sin }^{-1}x$
$1-x=\sin [\frac{\pi }{2}+2{\sin }^{-1}x]$
$1-x=\cos \left[2{\sin }^{-1}x\right]$
$1-x=1-2{\left[\sin \left({\sin }^{-1}x\right)\right]}^2[\because \cos \left(2x\right)=1-2{\sin }^{2}x]$
$1-x=1-2x^2$
$2x^2-x=0$
$x(2x-1)=0$
$x=0,x=\frac{1}{2}$
Put $x=0$ in equation (1)
LHS$={\sin }^{-1}1-2{\sin }^{-1}0=\frac{\pi }{2}=$RHS
Put $x=\frac{1}{2}$ in equation (1)
LHS$={\sin }^{-1}\left(\frac{1}{2}\right)-2{\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}-\frac{2\pi }{6}=-\frac{\pi }{6}$
LHS$\ne \frac{\pi }{2}\ne$RHS
So, $x=\frac{1}{2}$ is not a solution.
$\therefore x=0$ is a solution to given equation.