Solve: $\;$ $\cos^{-1} \left[\sin \left(\cos^{-1}x\right)\right] = \dfrac{\pi}{3}$
Given: $\;$ $\cos^{-1} \left[\sin \left(\cos^{-1}x\right)\right] = \dfrac{\pi}{3}$
i.e. $\;$ $\sin \left(\cos^{-1}x\right) = \cos \left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$ $\;\;\; \cdots \; (1)$
Let $\;$ $\cos^{-1}x = p$ $\;\;\; \cdots \; (2)$
$\therefore \;$ In view of equation $(2)$ equation $(1)$ becomes,
$\sin \left(p\right) = \dfrac{1}{2}$
i.e. $\;$ $p = \sin^{-1} \left(\dfrac{1}{2}\right) = \dfrac{\pi}{6}$
Substituting the value of $p$ in equation $(2)$, we get,
$\cos^{-1}x = \dfrac{\pi}{6}$
$\therefore \;$ $x = \cos \left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}$