The set of values of  x  for which the identity ${\cos }^{-1}x+{\cos }^{-1}(\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2})=\frac{\pi }{3}$holds good is

Case 1: If  $0\le x\le \frac{1}{2},$then

                        $\begin{array}{l}{\cos }^{-1}(x.\frac{1}{2}+\sqrt{1-x^2}\frac{\sqrt{3}}{2}) \text{= }{\cos }^{-1}x-{\cos }^{-1}\frac{1}{2}\\ \end{array}$

            Therefore, the equation is

            ${\cos }^{-1}x+{\cos }^{-1}x-{\cos }^{-1}\frac{1}{2}=\frac{\pi }{3}\Rightarrow x=\frac{1}{2}$

Case 2: If $\frac{1}{2}<x\le 1$,then 

            ${\cos }^{-1}(\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2})$$={\cos }^{-1}\frac{1}{2}-{\cos }^{-1}x$

            Therefore, the equation is 

            ${\cos }^{-1}x+{\cos }^{-1}\frac{1}{2}-{\cos }^{-1}x=\frac{\pi }{3},$

            Which is an identity.

            Hence, the equation holds good for $x\in [\frac{1}{2},1]$