Question

The solution of $Si{n}^{-1}x-Si{n}^{-1}$ $2x=\pm \frac{\pi }{3}$ 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)

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