Solve

$0\le \theta < 2\pi\\ \implies0\le \dfrac{2\theta}{3} <\dfrac{4\pi}{3}$

$\cot \left(\dfrac{2\theta}{3}\right)= -\sqrt3\\ \tan\left(\dfrac{2\theta}{3}\right) = -\dfrac{1}{\sqrt3}\\ \dfrac{2\theta}{3} = \dfrac{5\pi}{6}\\ \theta = \dfrac{5\pi}{4}$

$0\le\theta < 2\pi\\ \implies 0\le \dfrac{\theta}{2} < \pi\\ \implies \dfrac{\pi}{3} \le \dfrac{\theta}{2}+\dfrac{\pi}{3} < \dfrac{4\pi}{3}$

$\tan\left(\dfrac{\theta}{2}+\dfrac{\pi}{3}\right) = 1\\ \dfrac{\theta}{2}+\dfrac{\pi}{3} = \dfrac{5\pi}4\\ \theta = \dfrac{11\pi}{6}$