Question

The solution(s) of ${\sin }^{-1}x-{\sin }^{-1}2x=\pm \frac{\pi }{3}$ is

• A. $\pm \frac{1}{3}$
• B. $\pm \frac{1}{4}$
• C. $\pm \frac{1}{2}$
• D. $0$

Answer 1

The correct option is C $\pm \frac{1}{2}$

Given : ${\sin }^{-1}x-{\sin }^{-1}2x=\pm \frac{\pi }{3}$
Let ${\sin }^{-1}2x=\theta \Rightarrow 2x=\sin \theta$
We know that
$\theta \in [-\frac{\pi }{2},\frac{\pi }{2}],x\in [-\frac{1}{2},\frac{1}{2}]$

Now the given expression can be written as
${\sin }^{-1}\left(\frac{\sin \theta }{2}\right)-\theta =\pm \frac{\pi }{3}\Rightarrow {\sin }^{-1}\left(\frac{\sin \theta }{2}\right)=\theta \pm \frac{\pi }{3}\Rightarrow \frac{\sin \theta }{2}=\sin (\theta \pm \frac{\pi }{3})\Rightarrow \frac{\sin \theta }{2}=\frac{\sin \theta }{2}\pm \frac{\sqrt{3}\cos \theta }{2}\Rightarrow \cos \theta =0$
So, in the given interval the possible solutions are
$\theta =\pm \frac{\pi }{2}\Rightarrow 2x=\pm \sin \frac{\pi }{2}\therefore x=\pm \frac{1}{2}$