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

$=\frac{(2 \sin x-1)(\sin x-1)}{\sin x} $ 

$ \Rightarrow \mathop {\mathop {\sin x = \frac{1}{2}}\limits_ \Downarrow  }\limits_{4\,solutions} {\rm{ }}\,\,\,\,\,{\rm{or }}\,\,\,\,\mathop {\mathop {\,\frac{1}{2}{{\cos }^2}2x}\limits_{ \ge 0}  = \mathop {\sin x – 1}\limits_{ \le 0} }\limits_{{\rm{Hence no solution }}} $