sin(2x)<=1/2 inequation

Given the inequality:
$$\sin{\left(2 x \right)} \leq \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(2 x \right)} = \frac{1}{2}$$
Solve:
Given the equation
$$\sin{\left(2 x \right)} = \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = 2 \pi n + \operatorname{asin}{\left(\frac{1}{2} \right)}$$
$$2 x = 2 \pi n - \operatorname{asin}{\left(\frac{1}{2} \right)} + \pi$$
Or
$$2 x = 2 \pi n + \frac{\pi}{6}$$
$$2 x = 2 \pi n + \frac{5 \pi}{6}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \pi n + \frac{\pi}{12}$$
$$x_{2} = \pi n + \frac{5 \pi}{12}$$
$$x_{1} = \pi n + \frac{\pi}{12}$$
$$x_{2} = \pi n + \frac{5 \pi}{12}$$
This roots
$$x_{1} = \pi n + \frac{\pi}{12}$$
$$x_{2} = \pi n + \frac{5 \pi}{12}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{12}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{12}$$
substitute to the expression
$$\sin{\left(2 x \right)} \leq \frac{1}{2}$$
$$\sin{\left(2 \left(\pi n - \frac{1}{10} + \frac{\pi}{12}\right) \right)} \leq \frac{1}{2}$$

   /  1   pi         \       
sin|- - + -- + 2*pi*n| <= 1/2
   \  5   6          /       

one of the solutions of our inequality is:
$$x \leq \pi n + \frac{\pi}{12}$$

 _____           _____          
      \         /
-------•-------•-------
       x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \pi n + \frac{\pi}{12}$$
$$x \geq \pi n + \frac{5 \pi}{12}$$