Question

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

Answer 1

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

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

Let ${\sin }^{-1}\frac{x}{2}=\theta \to \frac{x}{2}=\sin \theta \to x=2\sin \theta$

$\Rightarrow {\cos }^{-1}x=\frac{\pi }{6}-\theta$

$x=\cos (\frac{\pi }{6}-\theta )$

$\Rightarrow x=\cos \frac{x}{6}\cos \theta +\sin \frac{\pi }{6}\sin \theta$

$x=\frac{\sqrt{3}}{2}\cos \theta +\frac{1}{2}\sin \theta$

$=\frac{\sqrt{3}}{2}\sqrt{1-{\sin }^2\theta }+\frac{1}{2}\sin \theta$

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

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

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

squaring both sides

$\Rightarrow \frac{9x^2}{16}=\frac{3}{16}(4-x^2)$

$9x^2=12-3x^2$

$12x^2=12$

$x^2=\pm 1$

$\because {\cos }^{-1}\frac{x}{4}+{\sin }^{-1}\frac{x}{2}=\frac{\pi }{6}$ so $x\ne -1$

$\therefore x=1$

Hence, the answer is $1.$