cost<-sqrt(3)/2 inequation

Given the inequality:
$$\cos{\left(t \right)} < \frac{\left(-1\right) \sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(t \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
Solve:
Given the equation
$$\cos{\left(t \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$t = \pi n + \operatorname{acos}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$t = \pi n - \pi + \operatorname{acos}{\left(- \frac{\sqrt{3}}{2} \right)}$$
Or
$$t = \pi n + \frac{5 \pi}{6}$$
$$t = \pi n - \frac{\pi}{6}$$
, where n - is a integer
$$t_{1} = \pi n + \frac{5 \pi}{6}$$
$$t_{2} = \pi n - \frac{\pi}{6}$$
$$t_{1} = \pi n + \frac{5 \pi}{6}$$
$$t_{2} = \pi n - \frac{\pi}{6}$$
This roots
$$t_{1} = \pi n + \frac{5 \pi}{6}$$
$$t_{2} = \pi n - \frac{\pi}{6}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$t_{0} < t_{1}$$
For example, let's take the point
$$t_{0} = t_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{5 \pi}{6}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{5 \pi}{6}$$
substitute to the expression
$$\cos{\left(t \right)} < \frac{\left(-1\right) \sqrt{3}}{2}$$
$$\cos{\left(\pi n - \frac{1}{10} + \frac{5 \pi}{6} \right)} < \frac{\left(-1\right) \sqrt{3}}{2}$$

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

but

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

Then
$$t < \pi n + \frac{5 \pi}{6}$$
no execute
one of the solutions of our inequality is:
$$t > \pi n + \frac{5 \pi}{6} \wedge t < \pi n - \frac{\pi}{6}$$

         _____  
        /     \  
-------ο-------ο-------
       t1      t2