cosx/(1+cos2x)<0 inequation

Given the inequality:
$$\frac{\cos{\left(x \right)}}{\cos{\left(2 x \right)} + 1} < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\cos{\left(x \right)}}{\cos{\left(2 x \right)} + 1} = 0$$
Solve:
Given the equation
$$\frac{\cos{\left(x \right)}}{\cos{\left(2 x \right)} + 1} = 0$$
transform
$$\frac{1}{2 \cos{\left(x \right)}} = 0$$
$$\frac{\cos{\left(x \right)}}{\cos{\left(2 x \right)} + 1} = 0$$
Do replacement
$$w = \cos{\left(2 x \right)}$$
Given the equation:
$$\frac{\cos{\left(x \right)}}{w + 1} = 0$$
Multiply the equation sides by the denominator 1 + w
we get:
$$\cos{\left(x \right)} = 0$$
Expand brackets in the left part

cosx = 0

This equation has no roots
do backward replacement
$$\cos{\left(2 x \right)} = w$$
Given the equation
$$\cos{\left(2 x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$2 x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
Or
$$2 x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$2 x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
substitute w:
This equation has no roots,
this inequality is executed for any x value or has no solutions
check it
subtitute random point x, for example

x0 = 0

$$\frac{\cos{\left(0 \right)}}{1 + \cos{\left(0 \cdot 2 \right)}} < 0$$

1/2 < 0

but

1/2 > 0

so the inequality has no solutions