sin2x>sqrt2/2 inequation

Given the inequality:
$$\sin{\left(2 x \right)} > \frac{\sqrt{2}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(2 x \right)} = \frac{\sqrt{2}}{2}$$
Solve:
Given the equation
$$\sin{\left(2 x \right)} = \frac{\sqrt{2}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = 2 \pi n + \operatorname{asin}{\left(\frac{\sqrt{2}}{2} \right)}$$
$$2 x = 2 \pi n - \operatorname{asin}{\left(\frac{\sqrt{2}}{2} \right)} + \pi$$
Or
$$2 x = 2 \pi n + \frac{\pi}{4}$$
$$2 x = 2 \pi n + \frac{3 \pi}{4}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \pi n + \frac{\pi}{8}$$
$$x_{2} = \pi n + \frac{3 \pi}{8}$$
$$x_{1} = \pi n + \frac{\pi}{8}$$
$$x_{2} = \pi n + \frac{3 \pi}{8}$$
This roots
$$x_{1} = \pi n + \frac{\pi}{8}$$
$$x_{2} = \pi n + \frac{3 \pi}{8}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{8}\right) - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{8}$$
substitute to the expression
$$\sin{\left(2 x \right)} > \frac{\sqrt{2}}{2}$$
$$\sin{\left(2 \left(\pi n - \frac{1}{10} + \frac{\pi}{8}\right) \right)} > \frac{\sqrt{2}}{2}$$

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   /1   pi\   \/ 2 
cos|- + --| > -----
   \5   4 /     2  
              

Then
$$x < \pi n + \frac{\pi}{8}$$
no execute
one of the solutions of our inequality is:
$$x > \pi n + \frac{\pi}{8} \wedge x < \pi n + \frac{3 \pi}{8}$$

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        /     \  
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       x_1      x_2