Question

Solve the following:

(i) sin⁻¹x + sin⁻¹2x = $\frac{\mathrm{\pi }}{3}$
(ii) ${\cos }^{-1}x+{\sin }^{-1}\frac{x}{2}=\frac{\mathrm{\pi }}{6}$

Answer 1

(i) We know
${\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}\left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right]$

$$\begin{gathered} \therefore {\sin }^{-1}x+{\sin }^{-1}2x=\frac{\mathrm{\pi }}{3} \\ \Rightarrow {\sin }^{-1}x+{\sin }^{-1}2x={\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ \Rightarrow {\sin }^{-1}x-{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=-{\sin }^{-1}2x \\ \Rightarrow {\sin }^{-1}\left[x\sqrt{1-\frac{3}{4}}+\frac{\sqrt{3}}{2}\sqrt{1-x^2}\right]=-{\sin }^{-1}2x \\ \Rightarrow {\sin }^{-1}\left[\frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-x^2}\right]={\sin }^{-1}\left(-2x\right) \\ \Rightarrow \frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-x^2}=-2x \\ \Rightarrow x+\sqrt{3}\sqrt{1-x^2}=-4x \\ \Rightarrow 5x=-\sqrt{3}\sqrt{1-x^2} \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}, \\ 25x^2=3-3x^2 \\ \Rightarrow 28x^2=3 \\ \Rightarrow x=\pm \frac{1}{2}\sqrt{\frac{3}{7}} \end{gathered}$$
(ii)

$$\begin{gathered} {\cos }^{-1}x+{\sin }^{-1}\frac{x}{2}=\frac{\mathrm{\pi }}{6} \\ \Rightarrow {\cos }^{-1}x+{\sin }^{-1}\frac{x}{2}={\sin }^{-1}\frac{1}{2} \\ \Rightarrow {\cos }^{-1}x={\sin }^{-1}\frac{1}{2}-{\sin }^{-1}\frac{x}{2} \\ \Rightarrow {\cos }^{-1}x={\sin }^{-1}\left[\frac{1}{2}\sqrt{1-\frac{x^2}{4}}-\frac{x}{2}\sqrt{1-\frac{1}{4}}\right]\left[\because {\sin }^{-1}x-{\sin }^{-1}y={\sin }^{-1}\left\{x\sqrt{1-y}-y\sqrt{1-x^2}\right\}\right] \\ \Rightarrow {\cos }^{-1}x={\sin }^{-1}\left[\frac{\sqrt{3}x}{4}-\frac{\sqrt{3}x}{4}\right] \\ \Rightarrow {\sin }^{-1}\sqrt{1-x^2}={\sin }^{-1}\left[\frac{\sqrt{3}x}{4}-\frac{\sqrt{3}x}{4}\right] \\ \Rightarrow \sqrt{1-x^2}=0 \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}, \\ \Rightarrow 1-x^2=0 \\ \Rightarrow x=\pm 1\left[\mathrm{As}x=-1\mathrm{is}\mathrm{not}\mathrm{satisfying}\mathrm{the}\mathrm{equation}\right] \end{gathered}$$