tg3x-sqrt3>=0 inequation

Given the inequality:
$$\tan{\left(3 x \right)} - \sqrt{3} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(3 x \right)} - \sqrt{3} = 0$$
Solve:
Given the equation
$$\tan{\left(3 x \right)} - \sqrt{3} = 0$$
- this is the simplest trigonometric equation

Move -sqrt(3) to right part of the equation

with the change of sign in -sqrt(3)

We get:
$$\tan{\left(3 x \right)} = \sqrt{3}$$
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
$$\tan{\left(3 x \right)} - \sqrt{3} \geq 0$$
$$\tan{\left(3 \left(\frac{\pi n}{3} - \frac{1}{10} + \frac{\pi}{9}\right) \right)} - \sqrt{3} \geq 0$$

    ___      /  3    pi       \     
- \/ 3  + tan|- -- + -- + pi*n| >= 0
             \  10   3        /     

but

    ___      /  3    pi       \    
- \/ 3  + tan|- -- + -- + pi*n| < 0
             \  10   3        /    

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

         _____  
        /
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       x1