Solve

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

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

$\Rightarrow-2 \sin ^{-1} x=\cos ^{-1}(1-x) \quad \ldots .(1)$

Let $\sin ^{-1} x=\theta \Rightarrow \sin \theta=x \Rightarrow \cos \theta=\sqrt{1-x^{2}}$.

$\therefore \theta=\cos ^{-1}\left(\sqrt{1-x^{2}}\right)$

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

Therefore, from equation (1), we have

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

Put $x=\sin y$. Then, we have:

$-2 \cos ^{-1}\left(\sqrt{1-\sin ^{2} y}\right)=\cos ^{-1}(1-\sin y)$

$\Rightarrow-2 \cos ^{-1}(\cos y)=\cos ^{-1}(1-\sin y)$

$\Rightarrow-2 y=\cos ^{-1}(1-\sin y)$

$\Rightarrow 1-\sin y=\cos (-2 y)=\cos 2 y$

$\Rightarrow 1-\sin y=1-2 \sin ^{2} y$

$\Rightarrow 2 \sin ^{2} y-\sin y=0$

$\Rightarrow \sin y(2 \sin y-1)=0$

$\Rightarrow \sin y=0$ or $\frac{1}{2}$

$\therefore x=0$ or $x=\frac{1}{2}$

But, when $x=\frac{1}{2}$, it can be observed that:

L.H.S. $=\sin ^{-1}\left(1-\frac{1}{2}\right)-2 \sin ^{-1} \frac{1}{2}$

$=\sin ^{-1}\left(\frac{1}{2}\right)-2 \sin ^{-1} \frac{1}{2}$

$=-\sin ^{-1} \frac{1}{2}$

$=-\frac{\pi}{6} \neq \frac{\pi}{2} \neq$ R.H.S.

$\therefore x=\frac{1}{2}$ is not the solution of the given equation.

Thus, $x=0$.

Hence, the correct answer is $\mathbf{C}$.