tan(2*x)>-sqrt(3) inequation

Given the inequality:
$$\tan{\left(2 x \right)} > - \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(2 x \right)} = - \sqrt{3}$$
Solve:
Given the equation
$$\tan{\left(2 x \right)} = - \sqrt{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = \pi n + \operatorname{atan}{\left(- \sqrt{3} \right)}$$
Or
$$2 x = \pi n - \frac{\pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} - \frac{\pi}{6}$$
$$x_{1} = \frac{\pi n}{2} - \frac{\pi}{6}$$
This roots
$$x_{1} = \frac{\pi n}{2} - \frac{\pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} - \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{\pi}{6} - \frac{1}{10}$$
substitute to the expression
$$\tan{\left(2 x \right)} > - \sqrt{3}$$
$$\tan{\left(2 \left(\frac{\pi n}{2} - \frac{\pi}{6} - \frac{1}{10}\right) \right)} > - \sqrt{3}$$

    /1   pi       \      ___
-tan|- + -- - pi*n| > -\/ 3 
    \5   3        /   

Then
$$x < \frac{\pi n}{2} - \frac{\pi}{6}$$
no execute
the solution of our inequality is:
$$x > \frac{\pi n}{2} - \frac{\pi}{6}$$

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       x1