cosx≤0 inequation

Given the inequality:
$$\cos{\left(x \right)} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(x \right)} = 0$$
Solve:
Given the equation
$$\cos{\left(x \right)} = 0$$
- this is the simplest trigonometric equation
We get:
$$\cos{\left(x \right)} = 0$$
This equation is transformed to
$$x = \pi n + \operatorname{acos}{\left(0 \right)}$$
$$x = \pi n - \pi + \operatorname{acos}{\left(0 \right)}$$
Or
$$x = \pi n + \frac{\pi}{2}$$
$$x = \pi n - \frac{\pi}{2}$$
, where n - is a integer
$$x_{1} = \pi n + \frac{\pi}{2}$$
$$x_{2} = \pi n - \frac{\pi}{2}$$
$$x_{1} = \pi n + \frac{\pi}{2}$$
$$x_{2} = \pi n - \frac{\pi}{2}$$
This roots
$$x_{1} = \pi n + \frac{\pi}{2}$$
$$x_{2} = \pi n - \frac{\pi}{2}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{2}\right) - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{2}$$
substitute to the expression
$$\cos{\left(x \right)} \leq 0$$
$$\cos{\left(\pi n - \frac{1}{10} + \frac{\pi}{2} \right)} \leq 0$$

    n               
(-1) *sin(1/10) <= 0
     

one of the solutions of our inequality is:
$$x \leq \pi n + \frac{\pi}{2}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \pi n + \frac{\pi}{2}$$
$$x \geq \pi n - \frac{\pi}{2}$$