cosx>(-√3)/2 inequation

Given the inequality:
$$\cos{\left(x \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(x \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
Solve:
Given the equation
$$\cos{\left(x \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
- 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
$$x = \pi n + \frac{5 \pi}{6}$$
$$x = \pi n - \frac{\pi}{6}$$
, 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:
$$x_{0} < x_{1}$$
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
$$\cos{\left(x \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$
$$\cos{\left(\pi n - \frac{1}{10} + \frac{5 \pi}{6} \right)} > \frac{\left(-1\right) \sqrt{3}}{2}$$

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

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

 _____           _____          
      \         /
-------ο-------ο-------
       x1      x2

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