tgx≥√3 inequation

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

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

but

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

Then
$$x \leq \pi n + \frac{\pi}{3}$$
no execute
the solution of our inequality is:
$$x \geq \pi n + \frac{\pi}{3}$$

         _____  
        /
-------•-------
       x1