sin2x>0 inequation

Given the inequality:
$$\sin{\left(2 x \right)} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(2 x \right)} = 0$$
Solve:
Given the equation
$$\sin{\left(2 x \right)} = 0$$
- this is the simplest trigonometric equation

with the change of sign in 0

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

sin(-1/5 + 2*pi*n) > 0

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

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       x1      x2