If ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$ then the vlaue of $x$ is 

The given equation is ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$

When $x=0$, the given equation is valid

Therefore, the solution for the equation ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$ is $x=\boxed{\mathbf{0}}$