cos(x)>=sqrt(3)/2 inequation

Given the inequality:
$$\cos{\left(x \right)} \geq \frac{\sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(x \right)} = \frac{\sqrt{3}}{2}$$
Solve:
Given the equation
$$\cos{\left(x \right)} = \frac{\sqrt{3}}{2}$$
- this is the simplest trigonometric equation
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
$$x = \pi n + \frac{\pi}{6}$$
$$x = \pi n - \frac{5 \pi}{6}$$
, where n - is a integer
$$x_{1} = \pi n + \frac{\pi}{6}$$
$$x_{2} = \pi n - \frac{5 \pi}{6}$$
$$x_{1} = \pi n + \frac{\pi}{6}$$
$$x_{2} = \pi n - \frac{5 \pi}{6}$$
This roots
$$x_{1} = \pi n + \frac{\pi}{6}$$
$$x_{2} = \pi n - \frac{5 \pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{6}\right) - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{6}$$
substitute to the expression
$$\cos{\left(x \right)} \geq \frac{\sqrt{3}}{2}$$
$$\cos{\left(\pi n - \frac{1}{10} + \frac{\pi}{6} \right)} \geq \frac{\sqrt{3}}{2}$$

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

but

                       ___
    n    /1    pi\   \/ 3 
(-1) *sin|-- + --| < -----
         \10   3 /     2  
                     

Then
$$x \leq \pi n + \frac{\pi}{6}$$
no execute
one of the solutions of our inequality is:
$$x \geq \pi n + \frac{\pi}{6} \wedge x \leq \pi n - \frac{5 \pi}{6}$$

         _____  
        /     \  
-------•-------•-------
       x_1      x_2