sin^2x>0 inequation

Given the inequality:
$$\sin^{2}{\left(x \right)} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin^{2}{\left(x \right)} = 0$$
Solve:
Given the equation:
$$\sin^{2}{\left(x \right)} = 0$$
Transform
$$\sin^{2}{\left(x \right)} = 0$$
$$- \left(\cos{\left(x \right)} - 1\right) \left(\cos{\left(x \right)} + 1\right) = 0$$
Consider each factor separately

Step

$$\cos{\left(x \right)} + 1 = 0$$
- this is the simplest trigonometric equation
Move $1$ to right part of the equation
with the change of sign in $1$
We get:
$$\cos{\left(x \right)} = -1$$
This equation is transformed to
$$x = 2 \pi n + \operatorname{acos}{\left(-1 \right)}$$
$$x = 2 \pi n - \pi + \operatorname{acos}{\left(-1 \right)}$$
Or
$$x = 2 \pi n + \pi$$
$$x = 2 \pi n$$
, where n - is a integer

Step

$$\cos{\left(x \right)} - 1 = 0$$
- this is the simplest trigonometric equation
Move $-1$ to right part of the equation
with the change of sign in $-1$
We get:
$$\cos{\left(x \right)} = 1$$
This equation is transformed to
$$x = 2 \pi n + \operatorname{acos}{\left(1 \right)}$$
$$x = 2 \pi n - \pi + \operatorname{acos}{\left(1 \right)}$$
Or
$$x = 2 \pi n$$
$$x = 2 \pi n - \pi$$
, where n - is a integer
The final answer:
$$x_{1} = 2 \pi n + \pi$$
$$x_{2} = 2 \pi n$$
$$x_{3} = 2 \pi n$$
$$x_{4} = 2 \pi n - \pi$$
$$x_{1} = 2 \pi n + \pi$$
$$x_{2} = 2 \pi n$$
$$x_{3} = 2 \pi n$$
$$x_{4} = 2 \pi n - \pi$$
$$x_{1} = 2 \pi n + \pi$$
$$x_{2} = 2 \pi n$$
$$x_{4} = 2 \pi n - \pi$$
This roots
$$x_{1} = 2 \pi n + \pi$$
$$x_{2} = 2 \pi n$$
$$x_{4} = 2 \pi n - \pi$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n + \pi\right) - \frac{1}{10}$$
=
$$2 \pi n - \frac{1}{10} + \pi$$
substitute to the expression
$$\sin^{2}{\left(x \right)} > 0$$
$$\sin^{2}{\left(2 \pi n - \frac{1}{10} + \pi \right)} > 0$$

   2          
sin (1/10) > 0
    

one of the solutions of our inequality is:
$$x < 2 \pi n + \pi$$

 _____           _____          
      \         /     \    
-------ο-------ο-------ο-------
       x_1      x_2      x_4

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < 2 \pi n + \pi$$
$$x > 2 \pi n \wedge x < 2 \pi n - \pi$$