tan^2x≤0 inequation

Given the inequality:
$$\tan^{2}{\left(x \right)} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan^{2}{\left(x \right)} = 0$$
Solve:
Given the equation
$$\tan^{2}{\left(x \right)} = 0$$
transform
$$\tan^{2}{\left(x \right)} = 0$$
$$\tan^{2}{\left(x \right)} = 0$$
Do replacement
$$w = \tan{\left(x \right)}$$
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 - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = 0$$
, then

D = b^2 - 4 * a * c = 

(0)^2 - 4 * (1) * (0) = 0

Because D = 0, then the equation has one root.

w = -b/2a = -0/2/(1)

$$w_{1} = 0$$
do backward replacement
$$\tan{\left(x \right)} = w$$
Given the equation
$$\tan{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{atan}{\left(w \right)}$$
Or
$$x = \pi n + \operatorname{atan}{\left(w \right)}$$
, where n - is a integer
substitute w:
$$x_{1} = \pi n + \operatorname{atan}{\left(w_{1} \right)}$$
$$x_{1} = \pi n + \operatorname{atan}{\left(0 \right)}$$
$$x_{1} = \pi n$$
$$x_{1} = 0$$
$$x_{1} = 0$$
This roots
$$x_{1} = 0$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10}$$
=
$$- \frac{1}{10}$$
substitute to the expression
$$\tan^{2}{\left(x \right)} \leq 0$$
$$\tan^{2}{\left(- \frac{1}{10} \right)} \leq 0$$

   2           
tan (1/10) <= 0
     

but

   2           
tan (1/10) >= 0
     

Then
$$x \leq 0$$
no execute
the solution of our inequality is:
$$x \geq 0$$

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