tg(x)<=-√3/3 inequation

Given the inequality:
$$\tan{\left(x \right)} \leq \frac{\left(-1\right) \sqrt{3}}{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(x \right)} = \frac{\left(-1\right) \sqrt{3}}{3}$$
Solve:
Given the equation
$$\tan{\left(x \right)} = \frac{\left(-1\right) \sqrt{3}}{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{atan}{\left(- \frac{\sqrt{3}}{3} \right)}$$
Or
$$x = \pi n - \frac{\pi}{6}$$
, where n - is a integer
$$x_{1} = \pi n - \frac{\pi}{6}$$
$$x_{1} = \pi n - \frac{\pi}{6}$$
This roots
$$x_{1} = \pi n - \frac{\pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n - \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{\pi}{6} - \frac{1}{10}$$
substitute to the expression
$$\tan{\left(x \right)} \leq \frac{\left(-1\right) \sqrt{3}}{3}$$
$$\tan{\left(\pi n - \frac{\pi}{6} - \frac{1}{10} \right)} \leq \frac{\left(-1\right) \sqrt{3}}{3}$$

                           ___ 
    /1    pi       \    -\/ 3  
-tan|-- + -- - pi*n| <= -------
    \10   6        /       3   
                        

the solution of our inequality is:
$$x \leq \pi n - \frac{\pi}{6}$$

 _____          
      \    
-------•-------
       x1