Solve $\tan 2x = -\cot \left(x + \frac{\pi}{3}\right)$

(N/A) We have,$\tan 2x = -\cot \left(x + \frac{\pi}{3}\right)$
Since $-\cot \theta = \tan \left(\frac{\pi}{2} + \theta\right)$,we get:
$\tan 2x = \tan \left(\frac{\pi}{2} + x + \frac{\pi}{3}\right)$
$\tan 2x = \tan \left(x + \frac{5\pi}{6}\right)$
The general solution for $\tan \theta = \tan \alpha$ is $\theta = n\pi + \alpha$,where $n \in \mathbb{Z}$.
Therefore,$2x = n\pi + x + \frac{5\pi}{6}$
$x = n\pi + \frac{5\pi}{6}$,where $n \in \mathbb{Z}$.