${\cos ^{ - 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right)$

As we know that the range of principal value branch of ${\cos ^{ - 1}}$ and ${\sin ^{ - 1}}$ are $[0,\;\pi ]$ and
$\left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right]$ respectively.

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

Then, $\frac{1}{2} = \cos \left( {\frac{\pi }{3}} \right),\;where\;\frac{\pi }{3} \in [0,\;\pi ]$

Let ${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = y \Rightarrow \frac{1}{2} = \sin y$

Then, $\frac{1}{2} = \sin \left( {\frac{\pi }{6}} \right),\;\;where\;\frac{\pi }{6} \in \left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right]$

$\therefore$ ${\cos ^{ - 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{3} + 2 \cdot \frac{\pi }{6} = \frac{\pi }{3} + \frac{\pi }{3} = \frac{{2\pi }}{3}.$