Number of roots of equation ${\sin }^{-1}(1-x)+2{\sin }^{-1}x=\frac{\pi }{2}$, is

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

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

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

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

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

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

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

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