Solve this

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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