Solve the following:

Solution:

(i) We know

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

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

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

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

$\Rightarrow \sin ^{-1}\left[x \sqrt{1-\frac{3}{4}}+\frac{\sqrt{3}}{2} \sqrt{1-x^{2}}\right]=-\sin ^{-1} 2 x$

$\Rightarrow \sin ^{-1}\left[\frac{x}{2}+\frac{\sqrt{3}}{2} \sqrt{1-x^{2}}\right]=\sin ^{-1}(-2 x)$

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

$\Rightarrow x+\sqrt{3} \sqrt{1-x^{2}}=-4 x$

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

Squaring both the sides,

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

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

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

(ii) 

$\cos ^{-1} x+\sin ^{-1} \frac{x}{2}=\frac{\pi}{6}$

$\Rightarrow \cos ^{-1} x+\sin ^{-1} \frac{x}{2}=\sin ^{-1} \frac{1}{2}$

$\Rightarrow \cos ^{-1} x=\sin ^{-1} \frac{1}{2}-\sin ^{-1} \frac{x}{2}$

$\Rightarrow \cos ^{-1} x=\sin ^{-1}\left[\frac{1}{2} \sqrt{1-\frac{x^{2}}{4}}-\frac{x}{2} \sqrt{1-\frac{1}{4}}\right] \quad\left[\because \sin ^{-1} x-\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y}-y \sqrt{1-x^{2}}\right\}\right]$

$\Rightarrow \cos ^{-1} x=\sin ^{-1}\left[\frac{\sqrt{3} x}{4}-\frac{\sqrt{3} x}{4}\right]$

$\Rightarrow \sin ^{-1} \sqrt{1-x^{2}}=\sin ^{-1}\left[\frac{\sqrt{3} x}{4}-\frac{\sqrt{3} x}{4}\right]$

$\Rightarrow \sqrt{1-x^{2}}=0$

Squaring both the sides,

$\Rightarrow 1-x^{2}=0$

$\Rightarrow x=\pm 1 \quad[$ As $x=-1$ is not satisfying the equation $]$