1-2cosx/2>0 inequation

Given the inequality:
$$- \frac{2 \cos{\left(x \right)}}{2} + 1 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- \frac{2 \cos{\left(x \right)}}{2} + 1 = 0$$
Solve:
Given the equation
$$- \frac{2 \cos{\left(x \right)}}{2} + 1 = 0$$
- this is the simplest trigonometric equation

Move 1 to right part of the equation

with the change of sign in 1

We get:
$$- \frac{2 \cos{\left(x \right)}}{2} = -1$$

Divide both parts of the equation by -1

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

1 - cos(-1/10 + pi*n) > 0

one of the solutions of our inequality is:
$$x < \pi n$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \pi n$$
$$x > \pi n - \pi$$