Trigonometric equation $\tan(\frac{\sqrt{3}x}{2})=-\sqrt{3}$ I want to solve a trigonometric equation below:
$$\tan(\frac{\sqrt{3}x}{2})=-\sqrt{3}$$
What is the value of $x$ for $x>0$ 
Thank you for your help.

Note $\tan\left(-\frac{\pi}{3}\right)=-\sqrt{3}$, and $\tan$ is periodic with period $\pi$, so this means that for some $n\in\mathbb{Z}$
$$\frac{\sqrt{3}x}{2}=n\pi-\frac{\pi}{3}\implies x=\frac{2\pi}{\sqrt{3}}\left(n-\frac{1}{3}\right).$$
Since we require $x>0$, this is equivalent to saying $n-\frac{1}{3}>0$, and since $n$ is an integer, we must have $n>0$. So in conclusion all possible solutions are of the form:
$$x=\frac{2\pi}{\sqrt{3}}\left(n-\frac{1}{3}\right),\quad n\in\mathbb{N}.$$
