Find out all solution of trigonometric equation $\tan \theta = - \frac{\sqrt 3}{3}$ $\tan \theta = - \frac{\sqrt 3}3$
I thought it was $2\pi\over3$ and $5\pi\over3$  but I was wrong
please help
 A: For any $a\in\Bbb R$,
$$\tan\theta=a\iff \theta=\arctan a+ n\pi,\, n\in\Bbb Z$$   
$\arctan x$ is the unique number $y$ in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ such that $\tan y=x$.
In this case, $$\tan\theta=-\frac{\sqrt{3}}{3}\iff \theta=-\frac{\pi}{6}+n\pi,\, n\in\Bbb Z,$$ 
because $\arctan\left(-\frac{\sqrt{3}}{3}\right)=-\frac{\pi}{6}$, which is in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and
$$\tan\left(-\frac{\pi}{6}\right)=\frac{\sin\left(-\frac{\pi}{6}\right)}{\cos\left(-\frac{\pi}{6}\right)}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$$   
There are infinitely many solutions for $\theta$, all generated by above formula.
A: $\tan \theta = - \frac{\sqrt 3}3=-\tan\frac{\pi}{6}$
$\implies \theta=n\pi+(-\frac{\pi}{6})~\quad where~ n\in \mathbb{Z}$
So there are infinitely many solution of given equation.
A: $$\tan \theta=-\frac{\sqrt{3}}{3}=-\frac{1}{\sqrt{3}}$$
$$\tan \theta=-\tan \left(\frac{\pi}{6}\right)=\tan \left(\frac{5\pi}{6}\right) $$
general solution:
$$\theta=n\pi+\frac{5\pi}{6}$$
