sin2x<(\sqrt(3))/(2) inequation

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

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

one of the solutions of our inequality is:
$$x < \pi n + \frac{\pi}{6}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \pi n + \frac{\pi}{6}$$
$$x > \pi n + \frac{\pi}{3}$$