The value of ${\cos }^{-1} x+{\cos }^{-1} \left(\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}\right)(1/2\le x\le 1)$ is equal to

Let ${\cos }^{-1} x=y.$ Then x = cos y, so that $1/2\le x\le 1\text{, or }0\le y\le \pi /3$, and

$\begin{array}{l}\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}=\frac{1}{2}\cos y+\frac{\sqrt{3}}{2}\sin y\\ =\cos \frac{\pi }{3}\cos y+\sin \frac{\pi }{3}\sin y=\cos \left(\frac{\pi }{3}-y\right)\\ \Rightarrow {\cos }^{-1} \left(\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}\right)=\frac{\pi }{3}-y\end{array}$

because ${\cos }^{-1} (\cos x)=x\text{ if }0\le x\le \pi$ Therefore, the given expression is equal to $y+\pi /3-y=\pi /3$.