$\sin \left(3x + \dfrac{\pi}{3}\right) - \cos \left(\dfrac{5\pi}{6} + 3x\right) = 2$
$\implies \sin \left(3x + \dfrac{\pi}{3}\right) - \cos \left(\dfrac{\pi}{2} - \left(-\dfrac{\pi}{3} - 3x\right)\right) = 2$
$\implies \sin \left(3x + \dfrac{\pi}{3}\right) - \sin \left(-\dfrac{\pi}{3} - 3x\right) = 2$
$\implies \sin \left(3x + \dfrac{\pi}{3}\right) + \sin \left(\dfrac{\pi}{3} + 3x\right) = 2$
$\implies 2\sin \left(3x + \dfrac{\pi}{3}\right) = 2$
$\implies \sin \left(3x + \dfrac{\pi}{3}\right) = 1$
$\implies 3x + \dfrac{\pi}{3} = \dfrac{\pi}{2} + k2\pi$
$\implies x = \dfrac{\pi}{18} + \dfrac{k2\pi}{3} \,\, (k \in Z)$