sin2x+sinx>0 inequation

Given the inequality:
$$\sin{\left(x \right)} + \sin{\left(2 x \right)} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(x \right)} + \sin{\left(2 x \right)} = 0$$
Solve:
$$x_{1} = 0$$
$$x_{2} = - \frac{4 \pi}{3}$$
$$x_{3} = - \pi$$
$$x_{4} = - \frac{2 \pi}{3}$$
$$x_{5} = \frac{2 \pi}{3}$$
$$x_{6} = \pi$$
$$x_{7} = \frac{4 \pi}{3}$$
$$x_{8} = 2 \pi$$
$$x_{1} = 0$$
$$x_{2} = - \frac{4 \pi}{3}$$
$$x_{3} = - \pi$$
$$x_{4} = - \frac{2 \pi}{3}$$
$$x_{5} = \frac{2 \pi}{3}$$
$$x_{6} = \pi$$
$$x_{7} = \frac{4 \pi}{3}$$
$$x_{8} = 2 \pi$$
This roots
$$x_{2} = - \frac{4 \pi}{3}$$
$$x_{3} = - \pi$$
$$x_{4} = - \frac{2 \pi}{3}$$
$$x_{1} = 0$$
$$x_{5} = \frac{2 \pi}{3}$$
$$x_{6} = \pi$$
$$x_{7} = \frac{4 \pi}{3}$$
$$x_{8} = 2 \pi$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{4 \pi}{3} - \frac{1}{10}$$
=
$$- \frac{4 \pi}{3} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(x \right)} + \sin{\left(2 x \right)} > 0$$
$$\sin{\left(2 \left(- \frac{4 \pi}{3} - \frac{1}{10}\right) \right)} + \sin{\left(- \frac{4 \pi}{3} - \frac{1}{10} \right)} > 0$$

     /1   pi\      /1    pi\    
- cos|- + --| + sin|-- + --| > 0
     \5   6 /      \10   3 /    

one of the solutions of our inequality is:
$$x < - \frac{4 \pi}{3}$$

 _____           _____           _____           _____           _____          
      \         /     \         /     \         /     \         /
-------ο-------ο-------ο-------ο-------ο-------ο-------ο-------ο-------
       x2      x3      x4      x1      x5      x6      x7      x8

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{4 \pi}{3}$$
$$x > - \pi \wedge x < - \frac{2 \pi}{3}$$
$$x > 0 \wedge x < \frac{2 \pi}{3}$$
$$x > \pi \wedge x < \frac{4 \pi}{3}$$
$$x > 2 \pi$$