Question:
$4 \sin ^{-1} x=\pi-\cos ^{-1} x$
Solution:
$4 \sin ^{-1} x=\pi-\cos ^{-1} x$
$\Rightarrow 4 \sin ^{-1} x=\pi-\left(\frac{\pi}{2}-\sin ^{-1} x\right)$ $\left[\because \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x\right]$
$\Rightarrow 4 \sin ^{-1} x=\frac{\pi}{2}+\sin ^{-1} x$
$\Rightarrow 3 \sin ^{-1} x=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{6}$
$\Rightarrow x=\sin \frac{\pi}{6}=\frac{1}{2}$
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