Solve this

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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