sin(x/6)<=-1/2 inequation

Given the inequality:
$$\sin{\left(\frac{x}{6} \right)} \leq - \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(\frac{x}{6} \right)} = - \frac{1}{2}$$
Solve:
Given the equation
$$\sin{\left(\frac{x}{6} \right)} = - \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{6} = 2 \pi n + \operatorname{asin}{\left(- \frac{1}{2} \right)}$$
$$\frac{x}{6} = 2 \pi n - \operatorname{asin}{\left(- \frac{1}{2} \right)} + \pi$$
Or
$$\frac{x}{6} = 2 \pi n - \frac{\pi}{6}$$
$$\frac{x}{6} = 2 \pi n + \frac{7 \pi}{6}$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{6}$$
$$x_{1} = 12 \pi n - \pi$$
$$x_{2} = 12 \pi n + 7 \pi$$
$$x_{1} = 12 \pi n - \pi$$
$$x_{2} = 12 \pi n + 7 \pi$$
This roots
$$x_{1} = 12 \pi n - \pi$$
$$x_{2} = 12 \pi n + 7 \pi$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(12 \pi n - \pi\right) + - \frac{1}{10}$$
=
$$12 \pi n - \pi - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(\frac{x}{6} \right)} \leq - \frac{1}{2}$$
$$\sin{\left(\frac{12 \pi n - \pi - \frac{1}{10}}{6} \right)} \leq - \frac{1}{2}$$

    /1    pi         \        
-sin|-- + -- - 2*pi*n| <= -1/2
    \60   6          /        

one of the solutions of our inequality is:
$$x \leq 12 \pi n - \pi$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq 12 \pi n - \pi$$
$$x \geq 12 \pi n + 7 \pi$$