cos3x – 2cos2x = 2 giải giúp ạ

$$\eqalign{
& \cos 3x – 2\cos 2x = 2 \cr
& \Leftrightarrow 4{\cos ^3}x – 3\cos x – 2\left( {2{{\cos }^2}x – 1} \right) = 2 \cr
& \Leftrightarrow 4{\cos ^3}x – 3\cos x – 4{\cos ^2}x + 2 = 2 \cr
& \Leftrightarrow 4{\cos ^3}x – 4{\cos ^2}x – 3\cos x = 0 \cr
& \Leftrightarrow \cos x\left( {4{{\cos }^2}x – 4\cos x – 3} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\cos x = 0 \hfill \cr
\cos x = {3 \over 2}\,\,\left( {loai} \right) \hfill \cr
\cos x = – {1 \over 2} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = {\pi \over 2} + k\pi \hfill \cr
x = \pm {{2\pi } \over 3} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$