-2sin(2x)<3^(1/2) inequation

Given the inequality:
$$- 2 \sin{\left(2 x \right)} < \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$- 2 \sin{\left(2 x \right)} = \sqrt{3}$$
Solve:
Given the equation
$$- 2 \sin{\left(2 x \right)} = \sqrt{3}$$
- this is the simplest trigonometric equation

Divide both parts of the equation by -2

The equation is transformed to
$$\sin{\left(2 x \right)} = - \frac{\sqrt{3}}{2}$$
This equation is transformed to
$$2 x = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$2 x = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$2 x = 2 \pi n - \frac{\pi}{3}$$
$$2 x = 2 \pi n + \frac{4 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \pi n - \frac{\pi}{6}$$
$$x_{2} = \pi n + \frac{2 \pi}{3}$$
$$x_{1} = \pi n - \frac{\pi}{6}$$
$$x_{2} = \pi n + \frac{2 \pi}{3}$$
This roots
$$x_{1} = \pi n - \frac{\pi}{6}$$
$$x_{2} = \pi n + \frac{2 \pi}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n - \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{\pi}{6} - \frac{1}{10}$$
substitute to the expression
$$- 2 \sin{\left(2 x \right)} < \sqrt{3}$$
$$- 2 \sin{\left(2 \left(\pi n - \frac{\pi}{6} - \frac{1}{10}\right) \right)} < \sqrt{3}$$

     /1   pi         \     ___
2*sin|- + -- - 2*pi*n| < \/ 3 
     \5   3          /   

but

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

Then
$$x < \pi n - \frac{\pi}{6}$$
no execute
one of the solutions of our inequality is:
$$x > \pi n - \frac{\pi}{6} \wedge x < \pi n + \frac{2 \pi}{3}$$

         _____  
        /     \  
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       x1      x2