Question

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

Answer 1

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

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

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

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

${\displaystyle \Rightarrow -{\sin }^{-1}x+{\sin }^{-1} \frac{x}{2}=\frac{2\pi -6\pi }{12}}$

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

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

${\displaystyle \Rightarrow \frac{x}{2}=\sin (-\frac{\pi }{3}+{\sin }^{-1}x)}$

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

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

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

${\displaystyle \Rightarrow 0=-\frac{\sqrt{3}}{2}\sqrt{1-x^2}}$

${\displaystyle \Rightarrow }$ 1 – x² = 0

${\displaystyle \Rightarrow }$ x² = 1

∴ x = 1 is the only answer because x = – 1 will not satisfy above question.