2*cos(2*x+pi/6)>sqrt(3) inequation

Given the inequality:
$$2 \cos{\left(2 x + \frac{\pi}{6} \right)} > \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \cos{\left(2 x + \frac{\pi}{6} \right)} = \sqrt{3}$$
Solve:
Given the equation
$$2 \cos{\left(2 x + \frac{\pi}{6} \right)} = \sqrt{3}$$
- this is the simplest trigonometric equation

Divide both parts of the equation by 2

The equation is transformed to
$$\cos{\left(2 x + \frac{\pi}{6} \right)} = \frac{\sqrt{3}}{2}$$
This equation is transformed to
$$2 x + \frac{\pi}{6} = \pi n + \operatorname{acos}{\left(\frac{\sqrt{3}}{2} \right)}$$
$$2 x + \frac{\pi}{6} = \pi n - \pi + \operatorname{acos}{\left(\frac{\sqrt{3}}{2} \right)}$$
Or
$$2 x + \frac{\pi}{6} = \pi n + \frac{\pi}{6}$$
$$2 x + \frac{\pi}{6} = \pi n - \frac{5 \pi}{6}$$
, where n - is a integer
Move
$$\frac{\pi}{6}$$
to right part of the equation
with the opposite sign, in total:
$$2 x = \pi n$$
$$2 x = \pi n - \pi$$
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{2}$$
$$x_{1} = \frac{\pi n}{2}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{2}$$
This roots
$$x_{1} = \frac{\pi n}{2}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{2}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\frac{\pi n}{2} + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{1}{10}$$
substitute to the expression
$$2 \cos{\left(2 x + \frac{\pi}{6} \right)} > \sqrt{3}$$
$$2 \cos{\left(2 \left(\frac{\pi n}{2} - \frac{1}{10}\right) + \frac{\pi}{6} \right)} > \sqrt{3}$$

     /  1   pi       \     ___
2*cos|- - + -- + pi*n| > \/ 3 
     \  5   6        /   

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

         _____  
        /     \  
-------ο-------ο-------
       x1      x2