2sinx+1>_0 inequation

Given the inequality:
$$2 \sin{\left(x \right)} + 1 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \sin{\left(x \right)} + 1 = 0$$
Solve:
Given the equation
$$2 \sin{\left(x \right)} + 1 = 0$$
- this is the simplest trigonometric equation

Move 1 to right part of the equation

with the change of sign in 1

We get:
$$2 \sin{\left(x \right)} = -1$$

Divide both parts of the equation by 2

The equation is transformed to
$$\sin{\left(x \right)} = - \frac{1}{2}$$
This equation is transformed to
$$x = 2 \pi n + \operatorname{asin}{\left(- \frac{1}{2} \right)}$$
$$x = 2 \pi n - \operatorname{asin}{\left(- \frac{1}{2} \right)} + \pi$$
Or
$$x = 2 \pi n - \frac{\pi}{6}$$
$$x = 2 \pi n + \frac{7 \pi}{6}$$
, where n - is a integer
$$x_{1} = 2 \pi n - \frac{\pi}{6}$$
$$x_{2} = 2 \pi n + \frac{7 \pi}{6}$$
$$x_{1} = 2 \pi n - \frac{\pi}{6}$$
$$x_{2} = 2 \pi n + \frac{7 \pi}{6}$$
This roots
$$x_{1} = 2 \pi n - \frac{\pi}{6}$$
$$x_{2} = 2 \pi n + \frac{7 \pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n - \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$2 \pi n - \frac{\pi}{6} - \frac{1}{10}$$
substitute to the expression
$$2 \sin{\left(x \right)} + 1 \geq 0$$
$$2 \sin{\left(2 \pi n - \frac{\pi}{6} - \frac{1}{10} \right)} + 1 \geq 0$$

         /1    pi         \     
1 - 2*sin|-- + -- - 2*pi*n| >= 0
         \10   6          /     

but

         /1    pi         \    
1 - 2*sin|-- + -- - 2*pi*n| < 0
         \10   6          /    

Then
$$x \leq 2 \pi n - \frac{\pi}{6}$$
no execute
one of the solutions of our inequality is:
$$x \geq 2 \pi n - \frac{\pi}{6} \wedge x \leq 2 \pi n + \frac{7 \pi}{6}$$

         _____  
        /     \  
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       x1      x2