Question

The solution of $Sin ^{-1}x -Sin ^{-1}$ $2x =\pm \frac {\pi}{3}$ is

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

Answer 1

Answer: (D)

$\sin ^{-1} \,x-\sin ^{-1} 2 x=\pm \frac{\pi}{3}$
$\Rightarrow \, \sin ^{-1} x-\sin ^{-1}\left(\pm \frac{\sqrt{3}}{2}\right)=\sin ^{-1} 2 x$
$\Rightarrow \sin ^{-1}\left[x \sqrt{1-\frac{3}{4}}-\left(\pm \frac{\sqrt{3}}{2} \sqrt{1-x^{2}}\right)\right]=\sin ^{-1} \,2 x$
$\Rightarrow \, \frac{x}{2}-\left(\pm \frac{\sqrt{3}}{2} \sqrt{1-x^{2}}\right)=2 x$
$\Rightarrow \,-\left(\pm \sqrt{3} \sqrt{1-x^{2}}\right)=3 x$
On squaring both sides, we get
$ 3\left(1-x^{2}\right)=9 x^{2}$
$ \Rightarrow \, 1-x^{2}=3 x^{2}$
$ \Rightarrow \,4 x^{2}=1$
$ \Rightarrow \, x=\pm \frac{1}{2}$