If $\cos^{-1} x = 2\sin^{-1} x$, what is $x$?

Explanation

1. 01

   $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$, so $\frac{\pi}{2} - \sin^{-1} x = 2\sin^{-1} x$ and $\sin^{-1} x = \frac{\pi}{6}$.

2. 02

   So $x = \sin \frac{\pi}{6} = \frac{1}{2}$; check: $\cos^{-1} \frac{1}{2} = \frac{\pi}{3} = 2 \times \frac{\pi}{6}$.