Question

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

Answer 1

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

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

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

$\Rightarrow \frac{x}{2}=\sin \theta$

$x=2\sin \theta$

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

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

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

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

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

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

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

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

squaring both sides

$\Rightarrow 9x^2=3(4-x^2)$

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

$12x^2=12$

$\Rightarrow x^2=\pm 1$

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

$\therefore x\ne -1$

$\Rightarrow x=1$

Hence, the answer is $1.$