${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\pi /2$. then x is equal to

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

$\begin{array}{l}\Rightarrow 1-x=\sin \left(\pi /2+2{\sin }^{-1}x\right)=\cos \left(2{\sin }^{-1}x\right)\\ =2\left(1-x^2\right)-1\\ \Rightarrow 2x^2-x=0\Rightarrow x=0\text{ or }1/2\end{array}$

But $x=1/2$ does not satisfy the given equation. Hence $x = 0.$