$$\eqalign{ & \cos 2x + 5 = 2\sqrt 2 \left( {2 - \cos x} \right)\sin \left( {x - {\pi \over 4}} \right) \cr & \Leftrightarrow {\cos ^2}x - {\sin ^2}x + 5 = 2\left( {2 - \cos x} \right)\left( {\sin x - \cos x} \right) \cr & \Leftrightarrow \left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) + 5 = 2\left( {2 - \cos x} \right)\left( {\sin x - \cos x} \right) \cr & \Leftrightarrow \left( {\cos x - \sin x} \right)\left( {\cos x + \sin x + 4 - 2\cos x} \right) + 5 = 0 \cr & \Leftrightarrow \left( {\cos x - \sin x} \right)\left( {\sin x - \cos x + 4} \right) + 5 = 0 \cr & \Leftrightarrow - {\left( {\cos x - \sin x} \right)^2} + 4\left( {\cos x - \sin x} \right) + 5 = 0 \cr & \Leftrightarrow \left[ \matrix{ \cos x - \sin x = - 1 \hfill \cr \cos x - \sin x = 5\,\,\left( {Vo\,\,nghiem} \right) \hfill \cr} \right. \cr & \Leftrightarrow \sqrt 2 \cos \left( {x + {\pi \over 4}} \right) = - 1 \cr & \Leftrightarrow \cos \left( {x + {\pi \over 4}} \right) = - {1 \over {\sqrt 2 }} \cr & \Leftrightarrow \left[ \matrix{ x + {\pi \over 4} = {{3\pi } \over 4} + k2\pi \hfill \cr x + {\pi \over 4} = - {{3\pi } \over 4} + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = {\pi \over 2} + k2\pi \hfill \cr x = - \pi + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$