Question

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

Answer 1

${\sin }^{-1}\left(\frac{x}{2}\right)=2$ ${\cos }^{-1}x=q$

$\frac{x}{2}=\sin P$ $\cos q=x$

$x=2sinP$ $\cos Q=2\sin p$....(2)

$p+q=\frac{\pi }{6}$

$P=\frac{\pi }{6}-Q$

$\sin p=\sin \frac{\pi }{6}-q)$

$=\sin \frac{\pi }{6}\cos Q-\cos \frac{\pi }{6}\sin q$

$=\frac{\cos q}{2}-\frac{\sqrt{3}}{2}\sin q$

$=2\sin p=(\cos q-\sqrt{3}\sin q)$....(1)

From (1) and (2)

$=\cos Q=\cos Q-\sqrt{3}\sin Q$

$\therefore \sin q=0\therefore q=o$

$\cos q=2\sin P$

$\cos \left(O\right)=2\sin P$

$\frac{1}{2}=\sin p\Rightarrow P=\frac{\pi }{6}$

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