Q. The number of solutions of the equation $x \sin ^{-1}(\sin x)+x=\pi$ in $[0,2 \pi]$ is
Inverse Trigonometric Functions
Solution:
$ x \sin ^{-1}(\sin x)+x=\pi$
$\sin ^{-1}(\sin x)=\frac{\pi}{x}-1$
$f(x)=\sin ^{-1} \sin x \text { and } g(x)=\frac{\pi}{x}-1$
