cos(2x)>sqrt(3)/2 inequation

Given the inequality:
$$\cos{\left(2 x \right)} > \frac{\sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(2 x \right)} = \frac{\sqrt{3}}{2}$$
Solve:
Given the equation
$$\cos{\left(2 x \right)} = \frac{\sqrt{3}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = \pi n + \operatorname{acos}{\left(\frac{\sqrt{3}}{2} \right)}$$
$$2 x = \pi n - \pi + \operatorname{acos}{\left(\frac{\sqrt{3}}{2} \right)}$$
Or
$$2 x = \pi n + \frac{\pi}{6}$$
$$2 x = \pi n - \frac{5 \pi}{6}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{12}$$
$$x_{2} = \frac{\pi n}{2} - \frac{5 \pi}{12}$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{12}$$
$$x_{2} = \frac{\pi n}{2} - \frac{5 \pi}{12}$$
This roots
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{12}$$
$$x_{2} = \frac{\pi n}{2} - \frac{5 \pi}{12}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} + \frac{\pi}{12}\right) - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{12}$$
substitute to the expression
$$\cos{\left(2 x \right)} > \frac{\sqrt{3}}{2}$$
$$\cos{\left(2 \left(\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{12}\right) \right)} > \frac{\sqrt{3}}{2}$$

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    n    /1   pi\   \/ 3 
(-1) *sin|- + --| > -----
         \5   3 /     2  
                    

Then
$$x < \frac{\pi n}{2} + \frac{\pi}{12}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{\pi n}{2} + \frac{\pi}{12} \wedge x < \frac{\pi n}{2} - \frac{5 \pi}{12}$$

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        /     \  
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       x_1      x_2