cos^2x-sin^2x>0 inequation

Given the inequality:
$$- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)} = 0$$
Solve:
Given the equation
$$- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)} = 0$$
transform
$$\cos{\left(2 x \right)} = 0$$
$$2 \cos^{2}{\left(x \right)} - 1 = 0$$
Do replacement
$$w = \cos{\left(x \right)}$$
This equation is of the form
$$a*w^2 + b*w + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$w_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$w_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 2$$
$$b = 0$$
$$c = -1$$
, then
$$D = b^2 - 4 * a * c = $$
$$0^{2} - 2 \cdot 4 \left(-1\right) = 8$$
Because D > 0, then the equation has two roots.
$$w_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$w_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$w_{1} = \frac{\sqrt{2}}{2}$$
Simplify
$$w_{2} = - \frac{\sqrt{2}}{2}$$
Simplify
do backward replacement
$$\cos{\left(x \right)} = w$$
$$\cos{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = 2 \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = 2 \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
Or
$$x = 2 \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = 2 \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, where n - is a integer
substitute w:
$$x_{1} = 2 \pi n + \operatorname{acos}{\left(w_{1} \right)}$$
$$x_{1} = 2 \pi n + \operatorname{acos}{\left(\frac{\sqrt{2}}{2} \right)}$$
$$x_{1} = 2 \pi n + \frac{\pi}{4}$$
$$x_{2} = 2 \pi n + \operatorname{acos}{\left(w_{2} \right)}$$
$$x_{2} = 2 \pi n + \operatorname{acos}{\left(- \frac{\sqrt{2}}{2} \right)}$$
$$x_{2} = 2 \pi n + \frac{3 \pi}{4}$$
$$x_{3} = 2 \pi n + \operatorname{acos}{\left(w_{1} \right)} - \pi$$
$$x_{3} = 2 \pi n - \pi + \operatorname{acos}{\left(\frac{\sqrt{2}}{2} \right)}$$
$$x_{3} = 2 \pi n - \frac{3 \pi}{4}$$
$$x_{4} = 2 \pi n + \operatorname{acos}{\left(w_{2} \right)} - \pi$$
$$x_{4} = 2 \pi n - \pi + \operatorname{acos}{\left(- \frac{\sqrt{2}}{2} \right)}$$
$$x_{4} = 2 \pi n - \frac{\pi}{4}$$
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{4}$$
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{4}$$
This roots
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{4}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{4} - \frac{1}{10}$$
=
$$- \frac{\pi}{4} - \frac{1}{10}$$
substitute to the expression
$$- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)} > 0$$
$$- \sin^{2}{\left(- \frac{\pi}{4} - \frac{1}{10} \right)} + \cos^{2}{\left(- \frac{\pi}{4} - \frac{1}{10} \right)} > 0$$

   2/1    pi\      2/1    pi\    
cos |-- + --| - sin |-- + --| > 0
    \10   4 /       \10   4 /    

Then
$$x < - \frac{\pi}{4}$$
no execute
one of the solutions of our inequality is:
$$x > - \frac{\pi}{4} \wedge x < \frac{\pi}{4}$$

         _____  
        /     \  
-------ο-------ο-------
       x_1      x_2