(a)[4]
If $\sin\left(x + \dfrac{\pi}{6}\right) - \sin\left(x - \dfrac{\pi}{6}\right) = \cos\left(x + \dfrac{\pi}{3}\right) - \cos\left(x - \dfrac{\pi}{3}\right)$, determine the exact value of $\tan x$.
(b)[2]
Hence determine the exact roots of the equation $\sin\left(x + \dfrac{\pi}{6}\right) - \sin\left(x - \dfrac{\pi}{6}\right) = \cos\left(x + \dfrac{\pi}{3}\right) - \cos\left(x - \dfrac{\pi}{3}\right)$ for $0 \le x \le 2\pi$.