2.$\frac{1}{2}$ (cos3x+ cos2x) + $\sqrt[]{3}$sin2x + cos3x=0
$\sqrt[]{3}$sin2x+ cos2x=-2cos3x
$\frac{\sqrt[]{3}}{2}$sin2x+$\frac{1}{2}$ cos2x=-cos3x
sin2x.cos $\frac{\pi}{6}$ +sin$\frac{\pi}{6}$.cos 2x=sin($\frac{-\pi}{2}$-3x)
sin(2x+$\frac{\pi}{6}$)=sin($\frac{-\pi}{2}$-3x)
2x+$\frac{\pi}{6}$=$\frac{-\pi}{2}$-3x+k2$\pi$ <-> x=$\frac{-2\pi}{15}$+$\frac{k2\pi}{5}$
hoặc 2x+$\frac{\pi}{6}$=$\pi$ +$\frac{\pi}{2}$+3x+k2$\pi$ <-> x=$\frac{-4\pi}{3}$+k2$\pi$
