Question

Solve:

${\sin }^{-1}x+{\sin }^{-1}2x=\frac{\pi }{3}$

Answer 1

Let $\theta =si{n}^{-1}x$ and $\varphi =si{n}^{-1}2x$

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

$\Rightarrow \theta +\varphi =\pi /3$

$\Rightarrow \varphi =\pi /3-\theta$

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

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

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

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

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

$\Rightarrow 28x^2=3$

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

$\Rightarrow x=\pm \sqrt{\frac{3}{28}}$