2*cos(x)+sqrt(3)>=0 inequation

Given the inequality:
$$2 \cos{\left(x \right)} + \sqrt{3} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \cos{\left(x \right)} + \sqrt{3} = 0$$
Solve:
Given the equation
$$2 \cos{\left(x \right)} + \sqrt{3} = 0$$
- this is the simplest trigonometric equation

Move sqrt(3) to right part of the equation

with the change of sign in sqrt(3)

We get:
$$2 \cos{\left(x \right)} = - \sqrt{3}$$

Divide both parts of the equation by 2

The equation is transformed to
$$\cos{\left(x \right)} = - \frac{\sqrt{3}}{2}$$
This equation is transformed to
$$x = \pi n + \operatorname{acos}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$x = \pi n - \pi + \operatorname{acos}{\left(- \frac{\sqrt{3}}{2} \right)}$$
Or
$$x = \pi n + \frac{5 \pi}{6}$$
$$x = \pi n - \frac{\pi}{6}$$
, where n - is a integer
$$x_{1} = \pi n + \frac{5 \pi}{6}$$
$$x_{2} = \pi n - \frac{\pi}{6}$$
$$x_{1} = \pi n + \frac{5 \pi}{6}$$
$$x_{2} = \pi n - \frac{\pi}{6}$$
This roots
$$x_{1} = \pi n + \frac{5 \pi}{6}$$
$$x_{2} = \pi n - \frac{\pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{5 \pi}{6}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{5 \pi}{6}$$
substitute to the expression
$$2 \cos{\left(x \right)} + \sqrt{3} \geq 0$$
$$2 \cos{\left(\pi n - \frac{1}{10} + \frac{5 \pi}{6} \right)} + \sqrt{3} \geq 0$$

  ___        /  1    pi       \     
\/ 3  - 2*sin|- -- + -- + pi*n| >= 0
             \  10   3        /     

one of the solutions of our inequality is:
$$x \leq \pi n + \frac{5 \pi}{6}$$

 _____           _____          
      \         /
-------•-------•-------
       x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \pi n + \frac{5 \pi}{6}$$
$$x \geq \pi n - \frac{\pi}{6}$$