Question

If $4{\cos }^{-1}x+{\sin }^{-1}x=\pi$, then the value of $x$ is.

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}$

$4{\cos }^{-1}x+{\sin }^{-1}x=\pi$

We know that ${\cos }^{-1}x+{\sin }^{-1}x=\frac{\pi }{2}$

$\Rightarrow {\cos }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x$

Now, $4{\cos }^{-1}x+{\sin }^{-1}x=\pi$

$\Rightarrow 4(\frac{\pi }{2}-{\sin }^{-1}x)+{\sin }^{-1}x=\pi$

$\Rightarrow 2\pi -4{\sin }^{-1}x+{\sin }^{-1}x=\pi$

$\Rightarrow -4{\sin }^{-1}x+{\sin }^{-1}x=\pi -2\pi$

$\Rightarrow -3{\sin }^{-1}x=-\pi$

$\Rightarrow 3{\sin }^{-1}x=\pi$

$\Rightarrow {\sin }^{-1}x=\frac{\pi }{3}$

$\Rightarrow x=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$

$\therefore x=\frac{\sqrt{3}}{2}$