Question

The soultion satisfying the equation $3 \cos ^{-1} x=\sin ^{-1}\left[\left(\sqrt{1-x^{2}}\right)\left(4 x^{2}-1\right)\right]$ is $[a, b]$ then $[ a + b ]$ equals ([.] denotes G.I.E)

Answer 1

$3 \cos ^{-1} x =\sin ^{-1}\left[\left(\sqrt{1- x ^{2}}\left(4 x ^{2}-1\right)\right)\right] $
put $ x = \cos \theta, 3 \cos ^{-1}(\cos \theta)$
$=\sin ^{-1}\left(\sin \theta\left(4 \cos ^{2} \theta-1\right)\right.$
$=\sin ^{-1}\left(\sin \theta\left(4\left(1-\sin ^{2} \theta\right)-1\right)\right.$
$=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right) $
$=\sin ^{-1}(\sin 3 \theta)$
equality holds when $0 \leq \theta \leq \pi^{\wedge}-\frac{\pi}{2} \leq 3 \theta \leq \frac{\pi}{2}$
$\Rightarrow 0 \leq \theta \leq \frac{\pi}{6} $
$\therefore \frac{\sqrt{3}}{2} \leq \cos \theta \leq 1$
$\therefore x \in\left[\frac{\sqrt{3}}{2}, 1\right]$