tg(2x)<-√3/3 inequation

Given the inequality:
$$\tan{\left(2 x \right)} < \frac{\left(-1\right) \sqrt{3}}{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(2 x \right)} = \frac{\left(-1\right) \sqrt{3}}{3}$$
Solve:
Given the equation
$$\tan{\left(2 x \right)} = \frac{\left(-1\right) \sqrt{3}}{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = \pi n + \operatorname{atan}{\left(- \frac{\sqrt{3}}{3} \right)}$$
Or
$$2 x = \pi n - \frac{\pi}{6}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} - \frac{\pi}{12}$$
$$x_{1} = \frac{\pi n}{2} - \frac{\pi}{12}$$
This roots
$$x_{1} = \frac{\pi n}{2} - \frac{\pi}{12}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} - \frac{\pi}{12}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{\pi}{12} - \frac{1}{10}$$
substitute to the expression
$$\tan{\left(2 x \right)} < \frac{\left(-1\right) \sqrt{3}}{3}$$
$$\tan{\left(2 \left(\frac{\pi n}{2} - \frac{\pi}{12} - \frac{1}{10}\right) \right)} < \frac{\left(-1\right) \sqrt{3}}{3}$$

                         ___ 
    /1   pi       \   -\/ 3  
-tan|- + -- - pi*n| < -------
    \5   6        /      3   
                      

the solution of our inequality is:
$$x < \frac{\pi n}{2} - \frac{\pi}{12}$$

 _____          
      \    
-------ο-------
       x1