$\begin{array}{l}\cos x+\sqrt3 \sin x = 2\cos 3x\\ \Leftrightarrow \dfrac{1}{2}\cos x + \dfrac{\sqrt3}{2}\sin x = \cos3x\\ \Leftrightarrow \cos\dfrac{\pi}{3}.\cos x + \sin\dfrac{\pi}{3}.\sin x = \cos3x\\ \Leftrightarrow \cos\left(\dfrac{\pi}{3}-x\right) = \cos3x\\ \Leftrightarrow \left[\begin{array}{l}\dfrac{\pi}{3} - x = 3x + k2\pi\\\dfrac{\pi}{3} - x = -3x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{9} + k\dfrac{2\pi}{3}\\x = -\dfrac{\pi}{6} + k\pi\end{array}\right.&(k \in \Bbb Z)\\ \\ 3\cos2x - 4\sin2x = 1\\ \Leftrightarrow 3(2\cos^2x -1) - 8\sin x\cos x = 1\\ \Leftrightarrow 3\cos^2x - 4\sin x\cos x = 2\\ \text{Nhận thấy cosx = 0 không là nghiệm của phương trình}\\ \text{Chia hai vế của phương trình cho $\cos^2x$, ta được:}\\ 3 - 4\tan x = \dfrac{2}{\cos^2x}\\ \Leftrightarrow 3 - 4\tan x = 2(1 + \tan^2x)\\ \Leftrightarrow 2\tan^2x + 4\tan x - 1 = 0\\ \Leftrightarrow \left[\begin{array}{l}\tan x = \dfrac{-2 - \sqrt6}{2}\\\tan x = \dfrac{-2 + \sqrt6}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \arctan\left(\dfrac{-2 - \sqrt6}{2}\right) + k\pi\\x = \arctan\left(\dfrac{-2 + \sqrt6}{2}\right) + k\pi\end{array}\right.&(k \in \Bbb Z)\\ \end{array}$
