Question

Let $S$ be the set of all solutions of the equation $\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^2}\right)=\pi$, $x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$. Then $\displaystyle\sum_{x \in S} 2 \sin ^{-1}\left(x^2-1\right)$ is equal to

• A. 0
• B. $\frac{-2 \pi}{3}$
• C. $\pi-2 \sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)$
• D. $\pi-\sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)$

Answer 1

Answer: (B)

$ \cos ^{-1}(2 x)-2 \cos ^{-1} \sqrt{1-x^2}=\pi $
$\cos ^{-1}(2 x)-\cos ^{-1}\left(2\left(1-x^2\right)-1\right)=\pi$
$ \cos ^{-1}(2 x)-\cos ^{-1}\left(1-2 x^2\right)=\pi $
$ -\cos ^{-1}\left(1-2 x^2\right)=\pi-\cos ^{-1}(2 x)$
Taking $\cos$ both sides we get
$\operatorname{Cos}\left(-\cos ^{-1}\left(1-2 x^2\right)\right)=\cos \left(\pi-\cos ^{-1}(2 x)\right)$
$ 1-2 x ^2=-2 x$
$2 x ^2-2 x -1=0$
On solving, $x =\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}$
As $x=[-1 / 2,1 / 2], x=\frac{1+\sqrt{3}}{2}=$ rejected
So $x=\frac{1-\sqrt{3}}{2} \Rightarrow x^2-1=-\sqrt{3} / 2$
$=2 \sin ^{-1}\left(x^2-1\right)=2 \sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{-2 \pi}{3}$