Question:
If $4 \cos ^{-1} x+\sin ^{-1} x=\pi$, then the value of $x$ is
(a) $\frac{3}{2}$
(b) $\frac{1}{\sqrt{2}}$
(c) $\frac{\sqrt{3}}{2}$
(d) $\frac{2}{\sqrt{3}}$
Solution:
(c) $\frac{\sqrt{3}}{2}$
We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$.
$4 \cos ^{-1} x+\sin ^{-1} x=\pi$
$\Rightarrow 4 \cos ^{-1} x+\frac{\pi}{2}-\cos ^{-1} x=\pi$
$\Rightarrow 3 \cos ^{-1} x=\pi-\frac{\pi}{2}$
$\Rightarrow 3 \cos ^{-1} x=\frac{\pi}{2}$
$\Rightarrow \cos ^{-1} x=\frac{\pi}{6}$
$\Rightarrow x=\cos \frac{\pi}{6}$
$\Rightarrow x=\frac{\sqrt{3}}{2}$
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