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

Answer

We know

$\sin^{-1}\text{x}+\sin^{-1}\text{y}=\sin^{-1}\Big[\text{x}\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big]$

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

$\Rightarrow\sin^{-1}\text{x}+\sin^{-1}2\text{x}=\sin^{-1}\Big(\frac{\sqrt{3}}{2}\Big)$

$\Rightarrow\sin^{-1}\text{x}-\sin^{-1}\Big(\frac{\sqrt{3}}{2}\Big)=-\sin^{-1}2\text{x}$

$\Rightarrow\sin^{-1}\bigg[\text{x}\sqrt{1-\frac{3}{4}}+\frac{\sqrt3}{2}\sqrt{1-\text{x}^2}\bigg]=-\sin^{-1}2\text{x}$

$\Rightarrow\sin^{-1}\Big[\frac{\text{x}}{2}+\frac{\sqrt3}{2}\sqrt{1-\text{x}^2}\Big]=\sin^{-1}(-2\text{x})$

$\Rightarrow\frac{\text{x}}{2}+\frac{\sqrt3}{2}\sqrt{1-\text{x}^2}=-2\text{x}$

$\Rightarrow5\text{x}=-\sqrt3\sqrt{1-\text{x}^2}$

Squaring both the sides,

$25\text{x}^2=3-3\text{x}^2$

$\Rightarrow28\text{x}^2=3$

$\Rightarrow\text{x}=\pm\frac{1}{2}\sqrt{\frac{3}{7}}$

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