Chia 2 vế cho 2 ta có
$\dfrac{1}{2} \sin(3x) + \dfrac{\sqrt{3}}{2} \cos(3x) = \dfrac{\sqrt{3}}{2}$
$<-> \sin(3x) \cos( \dfrac{\pi}{3}) + \sin(\dfrac{\pi}{3}) \cos(3x) = \sin( \dfrac{\pi}{3})$
$<-> \sin(3x + \dfrac{\pi}{3}) = \sin(\dfrac{\pi}{3})$
Vậy
$3x + \dfrac{\pi}{3} = \dfrac{\pi}{3} + 2k\pi$ hoặc $3x + \dfrac{\pi}{3} = \pi - \dfrac{\pi}{3} + 2k\pi$
Do đó
$x = \dfrac{2k\pi}{3}$ hoặc $x = \dfrac{\pi}{9} + \dfrac{2k\pi}{3}$.
