3tgx-√3>0 inequation

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

Divide both parts of the equation by 3

The equation is transformed to
$$\tan{\left(x \right)} = \frac{\sqrt{3}}{3}$$
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:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{6}\right) - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{6}$$
substitute to the expression
$$3 \tan{\left(x \right)} - \sqrt{3} > 0$$
$$3 \tan{\left(\pi n - \frac{1}{10} + \frac{\pi}{6} \right)} - \sqrt{3} > 0$$

    ___        /1    pi\    
- \/ 3  + 3*cot|-- + --| > 0
               \10   3 /    

Then
$$x < \pi n + \frac{\pi}{6}$$
no execute
the solution of our inequality is:
$$x > \pi n + \frac{\pi}{6}$$

         _____  
        /
-------ο-------
       x_1