sinx/2≥0 inequation

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

with the change of sign in 0

We get:
$$\frac{\sin{\left(x \right)}}{2} = 0$$

Divide both parts of the equation by 1/2

The equation is transformed to
$$\sin{\left(x \right)} = 0$$
This equation is transformed to
$$x = 2 \pi n + \operatorname{asin}{\left(0 \right)}$$
$$x = 2 \pi n - \operatorname{asin}{\left(0 \right)} + \pi$$
Or
$$x = 2 \pi n$$
$$x = 2 \pi n + \pi$$
, where n - is a integer
$$x_{1} = 2 \pi n$$
$$x_{2} = 2 \pi n + \pi$$
$$x_{1} = 2 \pi n$$
$$x_{2} = 2 \pi n + \pi$$
This roots
$$x_{1} = 2 \pi n$$
$$x_{2} = 2 \pi n + \pi$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$2 \pi n + - \frac{1}{10}$$
=
$$2 \pi n - \frac{1}{10}$$
substitute to the expression
$$\frac{\sin{\left(x \right)}}{2} \geq 0$$
$$\frac{\sin{\left(2 \pi n - \frac{1}{10} \right)}}{2} \geq 0$$

sin(-1/10 + 2*pi*n)     
------------------- >= 0
         2              

but

sin(-1/10 + 2*pi*n)    
------------------- < 0
         2             

Then
$$x \leq 2 \pi n$$
no execute
one of the solutions of our inequality is:
$$x \geq 2 \pi n \wedge x \leq 2 \pi n + \pi$$

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       x1      x2