giả nai hả em gái ? lần sau mà ko làm hết là anh xoá bài thẳng nhé he he
$\eqalign{
& co\;\left( {\sin x + \cos x} \right)\left( {\sin 2x + 2} \right) - 2\cos x\left( {{{\sin }^2}x + 1} \right) \cr
& = \sin x\sin 2x + 2\sin x + \cos x\sin 2x + 2\cos x - 2\cos x{\sin ^2}x - 2\cos x \cr
& = 2{\sin ^2}x\cos x + 2\sin x + \cos x\sin 2x - 2\cos x{\sin ^2}x \cr
& = 2\sin x + \cos x\sin 2x = 2\sin x + 2\sin x{\cos ^2}x \cr
& \to pt \Leftrightarrow 4\left( {1 - \cos x} \right) + \sin 2x\left( {2\sin x + 2\sin x{{\cos }^2}x} \right) = 0 \cr
& \Leftrightarrow 4\left( {1 - \cos x} \right) + 2{\sin ^2}x\cos x\left( {2 + 2{{\cos }^2}x} \right) = 0 \cr
& \Leftrightarrow 4\left( {1 - \cos x} \right) + \left( {1 - {{\cos }^2}x} \right)\left( {4\cos x + 4{{\cos }^3}x} \right) = 0 \cr
& \Leftrightarrow \left( {1 - \cos x} \right)\left[ {4 + \left( {1 + \cos x} \right)\left( {4\cos x + 4{{\cos }^3}x} \right)} \right] = 0 \cr
& TH1:\cos x = 1 \Leftrightarrow .... \cr
& TH2:4 + \left( {1 + \cos x} \right)\left( {4\cos x + 4{{\cos }^3}x} \right) = 0 \cr
& \Leftrightarrow {\cos ^4}x + {\cos ^3}x + {\cos ^2}x + \cos x + 1 = 0 \cr
& \Leftrightarrow {{{{\cos }^4}x} \over 2} + {{{{\left( {{{\cos }^2}x + \cos x} \right)}^2}} \over 2} + {{{{\left( {\cos x + 1} \right)}^2}} \over 2} + {1 \over 2} = 0 \cr
& vo\;nghiem \cr
& \Leftrightarrow ... \cr} $