sin(x/3)<=-1/(sqrt(2)) inequation

Given the inequality:
$$\sin{\left(\frac{x}{3} \right)} \leq - \frac{1}{\sqrt{2}}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(\frac{x}{3} \right)} = - \frac{1}{\sqrt{2}}$$
Solve:
Given the equation
$$\sin{\left(\frac{x}{3} \right)} = - \frac{1}{\sqrt{2}}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{3} = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{2}}{2} \right)}$$
$$\frac{x}{3} = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{2}}{2} \right)} + \pi$$
Or
$$\frac{x}{3} = 2 \pi n - \frac{\pi}{4}$$
$$\frac{x}{3} = 2 \pi n + \frac{5 \pi}{4}$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{3}$$
$$x_{1} = 6 \pi n - \frac{3 \pi}{4}$$
$$x_{2} = 6 \pi n + \frac{15 \pi}{4}$$
$$x_{1} = 6 \pi n - \frac{3 \pi}{4}$$
$$x_{2} = 6 \pi n + \frac{15 \pi}{4}$$
This roots
$$x_{1} = 6 \pi n - \frac{3 \pi}{4}$$
$$x_{2} = 6 \pi n + \frac{15 \pi}{4}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(6 \pi n - \frac{3 \pi}{4}\right) + - \frac{1}{10}$$
=
$$6 \pi n - \frac{3 \pi}{4} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(\frac{x}{3} \right)} \leq - \frac{1}{\sqrt{2}}$$
$$\sin{\left(\frac{6 \pi n - \frac{3 \pi}{4} - \frac{1}{10}}{3} \right)} \leq - \frac{1}{\sqrt{2}}$$

                             ___ 
    /1    pi         \    -\/ 2  
-sin|-- + -- - 2*pi*n| <= -------
    \30   4          /       2   
                          

one of the solutions of our inequality is:
$$x \leq 6 \pi n - \frac{3 \pi}{4}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq 6 \pi n - \frac{3 \pi}{4}$$
$$x \geq 6 \pi n + \frac{15 \pi}{4}$$