$$\eqalign{
& \left( {2\sin x - 1} \right)\left( {2\sin 2x + 1} \right) = 3 - 4{\cos ^2}x \cr
& \Leftrightarrow 4\sin x\sin 2x + 2\sin x - 2\sin 2x - 1 = 3 - 4{\cos ^2}x \cr
& \Leftrightarrow 8{\sin ^2}x\cos x + 2\sin x - 2\sin 2x - 4 + 4{\cos ^2}x = 0 \cr
& \Leftrightarrow 8{\sin ^2}x\cos x + 2\sin x - 2\sin 2x - 4\left( {1 - {{\cos }^2}x} \right) = 0 \cr
& \Leftrightarrow 8{\sin ^2}x\cos x + 2\sin x - 4\sin x\cos x - 4{\sin ^2}x = 0 \cr
& \Leftrightarrow 2\sin x\left( {4\sin x\cos x + 1 - 2\cos x - 2\sin x} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\sin x = 0\,\,\left( 1 \right) \hfill \cr
4\sin x\cos x + 1 - 2\sin x - 2\cos x = 0\,\,\,\left( 2 \right) \hfill \cr} \right. \cr
& \left( 1 \right) \Leftrightarrow x = k\pi \,\,\left( {k \in Z} \right) \cr
& \left( 2 \right) \Leftrightarrow 4\sin x\cos x - 2\left( {\sin x + \cos x} \right) + 1 = 0 \cr
& Dat\,\,t = \sin x + \cos x\,\,\left( { - \sqrt 2 \le t \le \sqrt 2 } \right) \cr
& \Rightarrow {t^2} = 1 + 2\sin x\cos x \cr
& \Leftrightarrow 2\sin x\cos x = {t^2} - 1 \cr
& PT:\,\,2\left( {{t^2} - 1} \right) - 2t + 1 = 0 \cr
& \Leftrightarrow 2{t^2} - 2t - 1 = 0 \Leftrightarrow \left[ \matrix{
t = {{1 + \sqrt 3 } \over 2}\, \hfill \cr
t = {{1 - \sqrt 3 } \over 2} \hfill \cr} \right.\,\,\left( {tm} \right) \cr
& t = {{1 \pm \sqrt 3 } \over 2}\, \Leftrightarrow \sqrt 2 \sin \left( {x + {\pi \over 4}} \right) = {{1 \pm \sqrt 3 } \over 2} \cr
& \Leftrightarrow \sin \left( {x + {\pi \over 4}} \right) = {{1 \pm \sqrt 3 } \over {2\sqrt 2 }} \cr
& \Leftrightarrow \left[ \matrix{
x + {\pi \over 4} = \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr
x + {\pi \over 4} = \pi - \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = - {\pi \over 4} + \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr
x = {{3\pi } \over 4} - \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$
