Solve for x: sin⁻¹(x/2) + cos⁻¹ x = π/6

Let $\theta = \sin^{-1}\frac{x}{2}$, so $x = 2\sin\theta$, $\cos\theta = \sqrt{1 - \frac{x^2}{4}}$. $\cos^{-1}x = \frac{\pi}{6} - \theta \Rightarrow x = \cos\left(\frac{\pi}{6} - \theta\right) = \frac{\sqrt3}{2}\cos\theta + \frac12\sin\theta$ $x = \frac{\sqrt3}{2}\sqrt{1 - \frac{x^2}{4}} + \frac{x}{4}$ $\frac{3x}{4} = \frac{\sqrt3}{4}\sqrt{4 - x^2}$, so $3x = \sqrt3\sqrt{4-x^2}$ Squaring: $9x^2 = 12 - 3x^2 \Rightarrow x^2 = 1$, so $x = \pm1$. Check $x = 1$: $\sin^{-1}\frac12 + \cos^{-1}1 = \frac{\pi}{6} + 0 = \frac{\pi}{6}$. True. Check $x = -1$: $\sin^{-1}\left(-\frac12\right) + \cos^{-1}(-1) = -\frac{\pi}{6} + \pi \neq \frac{\pi}{6}$. Rejected. Hence $x = 1$.