`cos(x-\frac{π}{3})=cos(2x+\frac{π}{6})`
⇔ $\left [\begin{array}{l} x-\dfrac{\pi}{3}=2x+\dfrac{\pi}{6}+k2π \\ x-\dfrac{\pi}{3}=-2x-\dfrac{\pi}{6}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} -x=\dfrac{\pi}{2}+k2π \\ 3x=\dfrac{\pi}{6}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=-\dfrac{\pi}{2}-k2π \\ x=\dfrac{\pi}{18}+\dfrac{k2π}{3} \end{array} \right. \ (k∈\mathbb{Z})$
