cos8x>-1 inequation

Given the inequality:
$$\cos{\left(8 x \right)} > -1$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(8 x \right)} = -1$$
Solve:
Given the equation
$$\cos{\left(8 x \right)} = -1$$
- this is the simplest trigonometric equation
This equation is transformed to
$$8 x = \pi n + \operatorname{acos}{\left(-1 \right)}$$
$$8 x = \pi n - \pi + \operatorname{acos}{\left(-1 \right)}$$
Or
$$8 x = \pi n + \pi$$
$$8 x = \pi n$$
, where n - is a integer
Divide both parts of the equation by
$$8$$
$$x_{1} = \frac{\pi n}{8} + \frac{\pi}{8}$$
$$x_{2} = \frac{\pi n}{8}$$
$$x_{1} = \frac{\pi n}{8} + \frac{\pi}{8}$$
$$x_{2} = \frac{\pi n}{8}$$
This roots
$$x_{1} = \frac{\pi n}{8} + \frac{\pi}{8}$$
$$x_{2} = \frac{\pi n}{8}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{8} + \frac{\pi}{8}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{8} - \frac{1}{10} + \frac{\pi}{8}$$
substitute to the expression
$$\cos{\left(8 x \right)} > -1$$
$$\cos{\left(8 \left(\frac{\pi n}{8} - \frac{1}{10} + \frac{\pi}{8}\right) \right)} > -1$$

-cos(-4/5 + pi*n) > -1

one of the solutions of our inequality is:
$$x < \frac{\pi n}{8} + \frac{\pi}{8}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{\pi n}{8} + \frac{\pi}{8}$$
$$x > \frac{\pi n}{8}$$