sin(10x)>=1/2 inequation

Given the inequality:
$$\sin{\left(10 x \right)} \geq \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(10 x \right)} = \frac{1}{2}$$
Solve:
Given the equation
$$\sin{\left(10 x \right)} = \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$10 x = 2 \pi n + \operatorname{asin}{\left(\frac{1}{2} \right)}$$
$$10 x = 2 \pi n - \operatorname{asin}{\left(\frac{1}{2} \right)} + \pi$$
Or
$$10 x = 2 \pi n + \frac{\pi}{6}$$
$$10 x = 2 \pi n + \frac{5 \pi}{6}$$
, where n - is a integer
Divide both parts of the equation by
$$10$$
get the intermediate answer:
$$x = \frac{\pi n}{5} + \frac{\pi}{60}$$
$$x = \frac{\pi n}{5} + \frac{\pi}{12}$$
$$x_{1} = \frac{\pi n}{5} + \frac{\pi}{60}$$
$$x_{2} = \frac{\pi n}{5} + \frac{\pi}{12}$$
$$x_{1} = \frac{\pi n}{5} + \frac{\pi}{60}$$
$$x_{2} = \frac{\pi n}{5} + \frac{\pi}{12}$$
This roots
$$x_{1} = \frac{\pi n}{5} + \frac{\pi}{60}$$
$$x_{2} = \frac{\pi n}{5} + \frac{\pi}{12}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{5} + \frac{\pi}{60}\right) - \frac{1}{10}$$
=
$$\frac{\pi n}{5} - \frac{1}{10} + \frac{\pi}{60}$$
substitute to the expression
$$\sin{\left(10 x \right)} \geq \frac{1}{2}$$
$$\sin{\left(10 \left(\frac{\pi n}{5} - \frac{1}{10} + \frac{\pi}{60}\right) \right)} \geq \frac{1}{2}$$

   /    pi\       
cos|1 + --| >= 1/2
   \    3 /       

but

   /    pi\      
cos|1 + --| < 1/2
   \    3 /      

Then
$$x \leq \frac{\pi n}{5} + \frac{\pi}{60}$$
no execute
one of the solutions of our inequality is:
$$x \geq \frac{\pi n}{5} + \frac{\pi}{60} \wedge x \leq \frac{\pi n}{5} + \frac{\pi}{12}$$

         _____  
        /     \  
-------•-------•-------
       x_1      x_2