$\eqalign{
& 2\sqrt 2 \cos 2x + \sin 2x\cos \left( {x + {{3\pi } \over 4}} \right) - 4\sin \left( {x + {\pi \over 4}} \right) = 0\;(1) \cr
& dk... \cr
& (1) \leftrightarrow 2\sqrt 2 \cos 2x + \sin 2x\left( {{{ - \sqrt 2 \cos x} \over 2} - {{\sqrt 2 \sin x} \over 2}} \right) - 4\left( {{{\sqrt 2 \sin x} \over 2} + {{\sqrt 2 \cos x} \over 2}} \right) = 0 \cr
& \leftrightarrow 4\cos 2x - \sin 2x\left( {\sin x + \cos x} \right) - 4\left( {\sin x + \cos x} \right) = 0 \cr
& \leftrightarrow 4\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) - \sin 2x\left( {\sin x + \cos x} \right) - 4\left( {\sin x + \cos x} \right) = 0 \cr
& \leftrightarrow \left( {\sin x + \cos x} \right)\left( {4\cos x - 4\sin x - \sin 2x - 4} \right) = 0 \cr
& TH1: \cr
& 4\cos x - 4\sin x - \sin 2x - 4 = 0 \cr
& dat\;t = cosx - sinx \cr
& \to {t^2} = {\cos ^2}x + \sin {x^2} - 2\sin x\cos x = 1 - \sin 2x \cr
& \to 4t + {t^2} - 5 = 0 \cr
& \leftrightarrow .... \cr
& TH2: \cr
& \sin x + \cos x = 0 \cr
& \leftrightarrow ... \cr
& \cr} $
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