$\begin{aligned} & \sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2} \\ & \Rightarrow \sin ^{-1}(1-x)=\left(\frac{\pi}{2}+2 \sin ^{-1} x\right) \\ & \Rightarrow 1-x=\sin \left(\frac{\pi}{2}+2 \sin ^{-1} x\right) \\ & \Rightarrow 1-x=\sin \frac{\pi}{2} \cos \left(2 \sin ^{-1} x\right)+\cos \frac{\pi}{2} \sin \left(2 \sin ^{-1} x\right) \\ & \Rightarrow 1-x=\cos \left(2 \sin ^{-1} x\right) \\ & \Rightarrow 1-x=\cos \left\{\cos ^{-1}\left(1-2 x\right)\right\} \Rightarrow 2 x^2-x=0 \\ & x=0, x=1 / 2 \text { which does not satisfy the equation } \\ & \therefore x=0 \text { is only the solution. }\end{aligned}$