The value of $${\cos ^{ - 1}}x + {\cos ^{ - 1}}\left( {\frac{x}{2} + \frac{1}{2}\sqrt {3 - 3{x^2}} } \right);\frac{1}{2} \leqslant x \leqslant 1{\text{ is}}$$

Solution :
$$\eqalign{ & {\text{Let, }}{\cos ^{ - 1}}x = y \cr & \Rightarrow x = \cos y,{\text{so that }}\frac{1}{2} \leqslant x \leqslant 1{\text{ or }}0 \leqslant y \leqslant \frac{\pi }{3} \cr & {\text{and }}\frac{x}{2} + \frac{1}{2}\sqrt {3 - 3{x^2}} = \frac{1}{2}\cos y + \frac{{\sqrt 3 }}{2}\sin y \cr & = \cos \frac{\pi }{3}\cos y + \sin \frac{\pi }{3}\sin y = \cos \left( {\frac{\pi }{3} - y} \right) \cr & \Rightarrow {\cos ^{ - 1}}\left( {\frac{x}{2} + \frac{1}{2}\sqrt {3 - 3{x^2}} } \right) = \frac{\pi }{3} - y \cr} $$
$$\therefore $$ the given expression is equal to
$$y + \frac{\pi }{3} - y,{\text{ i}}{\text{.e}}{\text{., }}\frac{\pi }{3}$$