Inequality with the tangent squared

Problem

Solve the inequality $\;\big[\tan\!\left(x-\tfrac{\pi}{3}\right)\big]^{2}\ge 3$.

Solution

Let $u=x-\dfrac{\pi}{3}$; the inequality is $\tan^{2}u\ge 3$, i.e. $|\tan u|\ge\sqrt3$, that is $\tan u\ge\sqrt3$ or $\tan u\le-\sqrt3$. Over one period: $u\in\Big[\tfrac{\pi}{3},\tfrac{\pi}{2}\Big)\cup\Big(\tfrac{\pi}{2},\tfrac{2\pi}{3}\Big],\qquad u\ne\tfrac{\pi}{2}.$ Returning to $x=u+\dfrac{\pi}{3}$: $\boxed{\,\tfrac{2\pi}{3}+k\pi\le x\le \pi+k\pi,\quad x\ne\tfrac{5\pi}{6}+k\pi\,}$