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Đáp án:

Giải thích các bước giải: Sửa 2cos^2x=2cos^2 2x

`2cos^2 2x -2cos2x +4sin6x +cos4x=1+4\sqrt{3}sin3x.cosx `

`⇔ 1-cos 4x -2cos2x +4sin6x +cos4x=1+4\sqrt{3}sin3x.cosx `

`⇔ 2(cos 4x-cos 2x)+8sin 3xcos 3x=4\sqrt{3}sin 3x.cos x`

`⇔ -4sin 3xsin x+8sin 3xcos 3x-4\sqrt{3}sin 3x.cos x=0`

`⇔ -4sin 3x(sin x-2cos 3x+\sqrt{3} cos x)=0`

`⇔` \(\left[ \begin{array}{l}-4sin 3x=0\\sin x-2cos 3x+\sqrt{3} cos x=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=k\dfrac{\pi}{3}\ (k \in \mathbb{Z})\\cos (x-\dfrac{\pi}{6})=cos 3x\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=k\dfrac{\pi}{3}\ (k \in \mathbb{Z})\\\left[ \begin{array}{l}x=-\dfrac{\pi}{12}+k\pi\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{24}+k\dfrac{\pi}{2}\ (k \in \mathbb{Z})\end{array} \right.\end{array} \right.\)