tgx>=sqrt(3)/3 inequation

Given the inequality:
$$\tan{\left(x \right)} \geq \frac{\sqrt{3}}{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(x \right)} = \frac{\sqrt{3}}{3}$$
Solve:
Given the equation
$$\tan{\left(x \right)} = \frac{\sqrt{3}}{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{atan}{\left(\frac{\sqrt{3}}{3} \right)}$$
Or
$$x = \pi n + \frac{\pi}{6}$$
, where n - is a integer
$$x_{1} = \pi n + \frac{\pi}{6}$$
$$x_{1} = \pi n + \frac{\pi}{6}$$
This roots
$$x_{1} = \pi n + \frac{\pi}{6}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{6}$$
substitute to the expression
$$\tan{\left(x \right)} \geq \frac{\sqrt{3}}{3}$$
$$\tan{\left(\pi n - \frac{1}{10} + \frac{\pi}{6} \right)} \geq \frac{\sqrt{3}}{3}$$

                           ___
   /  1    pi       \    \/ 3 
tan|- -- + -- + pi*n| >= -----
   \  10   6        /      3  
                         

but

                          ___
   /  1    pi       \   \/ 3 
tan|- -- + -- + pi*n| < -----
   \  10   6        /     3  
                        

Then
$$x \leq \pi n + \frac{\pi}{6}$$
no execute
the solution of our inequality is:
$$x \geq \pi n + \frac{\pi}{6}$$

         _____  
        /
-------•-------
       x1