All real $x$ satisfying $\sin^{-1}(x) + \sin^{-1}(1-x) = \cos^{-1}(x)$ are

Use $\cos^{-1}(x) = \tfrac{\pi}{2} - \sin^{-1}(x)$: equation becomes $\sin^{-1}(x) + \sin^{-1}(1-x) = \tfrac{\pi}{2} - \sin^{-1}(x)$ so $\sin^{-1}(1-x) = \tfrac{\pi}{2} - 2\sin^{-1}(x)$. **Take sine of both sides:** $1 - x = \sin\left(\tfrac{\pi}{2} - 2\sin^{-1}(x)\right) = \cos(2\sin^{-1}(x)) = 1 - 2x^2$. Therefore $1 - x = 1 - 2x^2 \Rightarrow 2x^2 - x = 0 \Rightarrow x(2x - 1) = 0$, giving $x = 0$ or $x = \tfrac{1}{2}$. Both satisfy the domain constraints $|x| \le 1$ and $|1-x| \le 1$, and both check in the original equation.