tg(x)+tg(2x)>=0 inequation

Given the inequality:
$$\tan{\left(x \right)} + \tan{\left(2 x \right)} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(x \right)} + \tan{\left(2 x \right)} = 0$$
Solve:
$$x_{1} = 0$$
$$x_{2} = - \frac{2 \pi}{3}$$
$$x_{3} = - \frac{\pi}{3}$$
$$x_{4} = \frac{\pi}{3}$$
$$x_{5} = \frac{2 \pi}{3}$$
$$x_{6} = \pi$$
$$x_{1} = 0$$
$$x_{2} = - \frac{2 \pi}{3}$$
$$x_{3} = - \frac{\pi}{3}$$
$$x_{4} = \frac{\pi}{3}$$
$$x_{5} = \frac{2 \pi}{3}$$
$$x_{6} = \pi$$
This roots
$$x_{2} = - \frac{2 \pi}{3}$$
$$x_{3} = - \frac{\pi}{3}$$
$$x_{1} = 0$$
$$x_{4} = \frac{\pi}{3}$$
$$x_{5} = \frac{2 \pi}{3}$$
$$x_{6} = \pi$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{2 \pi}{3} - \frac{1}{10}$$
=
$$- \frac{2 \pi}{3} - \frac{1}{10}$$
substitute to the expression
$$\tan{\left(x \right)} + \tan{\left(2 x \right)} \geq 0$$
$$\tan{\left(2 \left(- \frac{2 \pi}{3} - \frac{1}{10}\right) \right)} + \tan{\left(- \frac{2 \pi}{3} - \frac{1}{10} \right)} \geq 0$$

     /1   pi\      /1    pi\     
- tan|- + --| + cot|-- + --| >= 0
     \5   3 /      \10   6 /     

but

     /1   pi\      /1    pi\    
- tan|- + --| + cot|-- + --| < 0
     \5   3 /      \10   6 /    

Then
$$x \leq - \frac{2 \pi}{3}$$
no execute
one of the solutions of our inequality is:
$$x \geq - \frac{2 \pi}{3} \wedge x \leq - \frac{\pi}{3}$$

         _____           _____           _____  
        /     \         /     \         /     \  
-------•-------•-------•-------•-------•-------•-------
       x_2      x_3      x_1      x_4      x_5      x_6

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq - \frac{2 \pi}{3} \wedge x \leq - \frac{\pi}{3}$$
$$x \geq 0 \wedge x \leq \frac{\pi}{3}$$
$$x \geq \frac{2 \pi}{3} \wedge x \leq \pi$$