Question
Solve:
$\sin^{-1}\text{x}=\frac{\pi}{6}+\cos^{-1}\text{x}$
$\sin^{-1}\text{x}=\frac{\pi}{6}+\cos^{-1}\text{x}$
✓
Answer
$\frac{\pi}{2}-\cos^{-1}\text{x}=\frac{\pi}{6}+\cos^{-1}\text{x}$
$\frac{\pi}{3}=2\cos^{-1}\text{x}$
$\cos^{-1}\text{x}=\frac{\pi}{6}$
$\text{x}=\frac{\sqrt3}{2}$
$\frac{\pi}{3}=2\cos^{-1}\text{x}$
$\cos^{-1}\text{x}=\frac{\pi}{6}$
$\text{x}=\frac{\sqrt3}{2}$
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