Question

Solve the following :
sin${}^{-1}$ x + sin${}^{-1}$ $2$x = $\frac{\pi }{3}$

Answer 1

Let $\theta ={\sin }^{-1}\left(x\right)$ and $\varphi ={\sin }^{-1}\left(2x\right)$

Hence,

$\Rightarrow x=\sin \theta ,2x=\sin \varphi$

$\Rightarrow \varphi +\theta =\frac{\pi }{3}$

$\varphi =\frac{\pi }{3}-\theta$

$\Rightarrow \sin \varphi =\sin (\frac{\pi }{3}-\theta )$

$\sin \varphi =\sin \frac{\pi }{3}\cos \theta -\cos \frac{\pi }{3}\sin \theta$

$2x=\frac{\sqrt{3}}{2}\sqrt{1-x^2}-\frac{x}{2}$

$\frac{5x}{2}=\frac{\sqrt{3}\times \sqrt{1-x^2}}{2}$

Squaring both sides,

$\Rightarrow 25x^2=3(1-x^2)$

$25x^2=3-3x^2$

$x^2=\frac{3}{28}$

$\therefore x=\frac{\sqrt{3}}{\sqrt{28}}$

Hence, solved.