Official solution
\[{\cos ^{ - 1}}x - 2{\sin ^{ - 1}}x = {\cos ^{ - 1}}2x\]
For Domain : \[x \in \left[ {{{ - 1} \over 2},{1 \over 2}} \right]\]
\[{\cos ^{ - 1}}x - 2\left( {{\pi \over 2} - {{\cos }^{ - 1}}x} \right) = {\cos ^{ - 1}}(2x)\]
\[ \Rightarrow {\cos ^{ - 1}}x + 2{\cos ^{ - 1}}x = \pi + {\cos ^{ - 1}}2x\]
\[ \Rightarrow \cos (3{\cos ^{ - 1}}x) = - \cos ({\cos ^{ - 1}}2x)\]
\[ \Rightarrow 4{x^3} = x\]
\[ \Rightarrow x = 3,\, \pm \,{1 \over 2}\]