Solve the equation for x: sin⁻¹ x + sin⁻¹(1 - x) = cos⁻¹ x, x ≠ 0

Using $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$: $\sin^{-1}(1-x) = \frac{\pi}{2} - 2\sin^{-1}x$ Taking sine of both sides: $1 - x = \cos(2\sin^{-1}x) = 1 - 2x^2$ $2x^2 - x = 0 \Rightarrow x(2x - 1) = 0$ Since $x \neq 0$, $x = \frac{1}{2}$. Check: $\sin^{-1}\frac12 + \sin^{-1}\frac12 = \frac{\pi}{3} = \cos^{-1}\frac12$. Hence $x = \frac{1}{2}$.