Question
Solve:
$4\sin^{-1}\text{x}={\pi}-\cos^{-1}\text{x}$

Answer

$4\sin^{-1}\text{x}={\pi}-\cos^{-1}\text{x}$
$\Rightarrow4\sin^{-1}\text{x}={\pi}-\Big(\frac{\pi}{2}-\sin^{-1}\text{x}\Big)$
$\Big[\because\ \cos^{-1}\text{x}=\frac{\pi}{2}-\sin^{-1}\text{x}\Big]$
$\Rightarrow4\sin^{-1}\text{x}=\frac{\pi}{2}+\sin^{-1}\text{x}$
$\Rightarrow3\sin^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\sin^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\sin\frac{\pi}{6}=\frac{1}{2}$

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