We have,
$\begin{aligned}
& \sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x \\
& \Rightarrow \sin ^{-1}(1-x)=\frac{\pi}{2}-\sin ^{-1} x-\sin ^{-1} x \\
& =\frac{\pi}{2}-2 \sin ^{-1} x \\
& {\left[\because \cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}\right]} \\
& \Rightarrow \sin ^{-1}(1-x)=\sin ^{-1} 1-\sin ^{-1} 2 x \sqrt{1-x^2} \\
& \Rightarrow \sin ^{-1}(1-x)=\sin ^{-1}\left(1 \sqrt{1-4 x^2\left(1-x^2\right)}-0\right) \\
& \Rightarrow \quad(1-x)=\sqrt{1-4 x^4+4 x^2}=\sqrt{\left(1-2 x^2\right)^2} \\
& \Rightarrow \quad(1-x)=1-2 x^2 \\
& \Rightarrow \quad 2 x^2-x=0 \\
& \Rightarrow \quad x(2 x-1)=0 \\
& \Rightarrow \quad x=0, \frac{1}{2} \\
& \therefore \quad x \in\left\{0, \frac{1}{2}\right\} \\
&
\end{aligned}$