Expand each term using \(\sin(u\pm v)=\sin u\cos v\pm\cos u\sin v\) and \(\cos(u\pm v)=\cos u\cos v\mp\sin u\sin v\):

\[ \sin\!\left(x+\tfrac{\pi}{6}\right) -\sin\!\left(x-\tfrac{\pi}{6}\right) =(\sin x\cos\tfrac{\pi}{6}+\cos x\sin\tfrac{\pi}{6}) -(\sin x\cos\tfrac{\pi}{6}-\cos x\sin\tfrac{\pi}{6}) \] \[ =2\cos x\sin\tfrac{\pi}{6} =2\cos x\cdot\tfrac12 =\cos x. \]

\[ \cos\!\left(x+\tfrac{\pi}{3}\right) -\cos\!\left(x-\tfrac{\pi}{3}\right) =(\cos x\cos\tfrac{\pi}{3}-\sin x\sin\tfrac{\pi}{3}) -(\cos x\cos\tfrac{\pi}{3}+\sin x\sin\tfrac{\pi}{3}) \] \[ =-2\sin x\sin\tfrac{\pi}{3} =-2\sin x\cdot\tfrac{\sqrt3}{2} =-\sqrt3\,\sin x. \]

Equate LHS and RHS: \[ \cos x=-\sqrt3\,\sin x \quad\Rightarrow\quad \tan x=\frac{\sin x}{\cos x}=-\frac{1}{\sqrt3}. \]

(b) Since \(\tan x = -\tfrac{1}{\sqrt3}\) (reference angle \(=\tfrac{\pi}{6}\)) and tangent is negative in quadrants II and IV, \[ x = \pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}, \qquad x = 2\pi-\tfrac{\pi}{6}=\tfrac{11\pi}{6}. \]

(a) \(\displaystyle \tan x=-\frac{1}{\sqrt3}\).
(b) \(\displaystyle x=\frac{5\pi}{6},\;\frac{11\pi}{6}\) on \(0\le x\le 2\pi\).