Question

Solve the following:

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

Answer 1

${\displaystyle {\cos }^{-1}x+{\sin }^{-1}\left(\frac{x}{2}\right)=\frac{\pi }{6}}$

${\displaystyle (\frac{\pi }{2}-{\sin }^{-1}x)+{\sin }^{-1}\left(\frac{x}{2}\right)=\frac{\pi }{6}}$

${\displaystyle \frac{\pi }{2}-\frac{\pi }{6}={\sin }^{-1}x-{\sin }^{-1}\left(\frac{x}{2}\right)}$

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

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

${\displaystyle {\sin }^{-1}\left(x\right)={\sin }^{-1}(\frac{\sqrt{3}}{2}\sqrt{1-\frac{x^2}{4}}+\frac{x}{2}\sqrt{1-\frac{3}{4}}) ...[{\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1} [x\sqrt{1-y^2}+y\sqrt{1-x^2}].}$

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

${\displaystyle x=\frac{\sqrt{3}\sqrt{4-x^2}}{4}+\frac{x}{4}}$

${\displaystyle x-\frac{x}{4}=\frac{\sqrt{3}\sqrt{4-x^2}}{4}}$

${\displaystyle \frac{3x}{\cancel{4}}= \frac{\sqrt{3}\sqrt{4-x^2}}{\cancel{4}}}$

Squaring both the sides

9x² = 3(4 − x²)

3x² = 4 - x²

3x² + x² = 4

4x² = 4

x² = 1

x = ± 1.