⇔12cos4x−√32.sin4x+2.sin2x.\cosx=12${\Leftrightarrow} {\frac{1}{2}} (1 - 2. {\sin^{2}2x}) + {\sqrt{3}}. {\sin 2x}.{\cos2x} + 2.{\sin2x}.{\cosx} = {\frac{1}{2}}$
${\Leftrightarrow} - {\sin^{2}2x}) + {\sqrt{3}}. {\sin 2x}.{\cos2x} + 2.{\sin2x}.{\cosx} = 0$
${\Leftrightarrow} -{\sin2x}.( {\sin 2x} - {\sqrt{3}}. {\cos2x} - 2.{\cosx}) = 0$
\displaystyle{{\Leftrightarrow} -2{\sin2x}.({\sin(2x - {\Pi/6}) - {\cosx}) = 0}
ta dc: ${\sin2x} =0 va {\sin(2x - {\Pi/6}) - {\cosx} =0Leftrightarrow} {\sin(2x - {\Pi/6})= {\sin({\Pi/2} -x)}=0$