Let $\sin^{-1}(\large\frac{1}{2}) = x$ $ \Rightarrow \sin x = \large\frac{1}{2}$
$\Rightarrow \sin x = \large\frac{1}{x} = \sin \large\frac{\pi}{6}$
\( \therefore\) $x =\large\frac{\pi}{6} \; \epsilon \; \left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
Let $\cos^{-1}(\large\frac{1}{2}) = y\: $ $ \Rightarrow\: \cos y = \large\frac{1}{2}$
$\Rightarrow \cos y = \large\frac{1}{2} = \cos \large\frac{\pi}{3}$
\( \therefore\) $y = \large\frac{\pi}{3} \; \epsilon \; \left [ 0,\pi \right ]$
$\cos^{-1}(\large\frac{1}{2}) + 2 \sin^{-1}(\large\frac{1}{2}) = y + 2x$
$\Rightarrow \large\frac{\pi}{3} + 2\large\frac{\pi}{6} = \large\frac{\pi}{3} + \large\frac{\pi}{3} = 2 \large\frac{\pi}{3} $
\( \therefore\) $\cos^{-1}(\large\frac{1}{2}) + 2 \sin^{-1}(\large\frac{1}{2}) = 2 \large\frac{\pi}{3} $