[TEX]2\cos 6x +2\cos 4x = 2\sin x\cos x +\sqrt 3\ (1+\cos 2x)[/TEX]
[TEX]\Leftrightarrow 2(\cos 6x +\cos 4x) = 2\sin x\cos x+2\sqrt 3 \cos^2 x[/TEX]
[TEX]\Leftrightarrow 2\cos 5x\cos x = \cos x (\sin x +\sqrt 3\cos x)[/TEX]
[TEX]\Leftrightarrow \cos x(\sin x + \sqrt 3\cos x - 2\cos 5x)=0[/TEX]
[TEX]\cos x=0 (1)[/TEX] hoặc [TEX]\sin x +\sqrt 3 \cos x = 2\cos 5x (2)[/TEX]
[TEX](1): x= \frac{\pi}{2}+k\pi[/TEX]
[TEX](2): \frac{1}{2}\sin x+\frac{\sqrt 3}{2}\cos x = \cos 5x[/TEX]
[TEX]\Leftrightarrow \cos (x-\frac{\pi}{6}) =\cos 5x[/TEX]
Đến đây bạn tự giải nốt nhé