tg(3x)>=sqrt(3) inequation

Given the inequality:
$$\tan{\left(3 x \right)} \geq \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(3 x \right)} = \sqrt{3}$$
Solve:
Given the equation
$$\tan{\left(3 x \right)} = \sqrt{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$3 x = \pi n + \operatorname{atan}{\left(\sqrt{3} \right)}$$
Or
$$3 x = \pi n + \frac{\pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$3$$
$$x_{1} = \frac{\pi n}{3} + \frac{\pi}{9}$$
$$x_{1} = \frac{\pi n}{3} + \frac{\pi}{9}$$
This roots
$$x_{1} = \frac{\pi n}{3} + \frac{\pi}{9}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{3} + \frac{\pi}{9}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{3} - \frac{1}{10} + \frac{\pi}{9}$$
substitute to the expression
$$\tan{\left(3 x \right)} \geq \sqrt{3}$$
$$\tan{\left(3 \left(\frac{\pi n}{3} - \frac{1}{10} + \frac{\pi}{9}\right) \right)} \geq \sqrt{3}$$

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

but

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

Then
$$x \leq \frac{\pi n}{3} + \frac{\pi}{9}$$
no execute
the solution of our inequality is:
$$x \geq \frac{\pi n}{3} + \frac{\pi}{9}$$

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