Solve: sin⁻¹(x) + sin⁻¹(1-x) = cos⁻¹ x. (Application)

Use $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$: $\sin^{-1}(1-x) = \frac{\pi}{2} - 2\sin^{-1}x$. Taking sine: $1 - x = \cos(2\sin^{-1}x) = 1 - 2x^2$, so $2x^2 - x = 0$ and $x = 0$ or $x = \frac{1}{2}$. Both satisfy the equation (for $x = \frac12$: $\frac{\pi}{6}+\frac{\pi}{6} = \frac{\pi}{3} = \cos^{-1}\frac12$; for $x = 0$: $0 + \frac{\pi}{2} = \cos^{-1}0$). Ans: $x = 0, \frac{1}{2}$