Given the inequality:
$$\cot^{2}{\left(x \right)} + \cot{\left(x \right)} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot^{2}{\left(x \right)} + \cot{\left(x \right)} = 0$$
Solve:
Given the equation
$$\cot^{2}{\left(x \right)} + \cot{\left(x \right)} = 0$$
transform
$$\left(\cot{\left(x \right)} + 1\right) \cot{\left(x \right)} = 0$$
$$\cot^{2}{\left(x \right)} + \cot{\left(x \right)} = 0$$
Do replacement
$$w = \cot{\left(x \right)}$$
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
$$a = 1$$
$$b = 1$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (1) * (0) = 1
Because D > 0, then the equation has two roots.
w1 = (-b + sqrt(D)) / (2*a)
w2 = (-b - sqrt(D)) / (2*a)
or
$$w_{1} = 0$$
$$w_{2} = -1$$
do backward replacement
$$\cot{\left(x \right)} = w$$
substitute w:
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{2}$$
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{2}$$
This roots
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{2}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{4} - \frac{1}{10}$$
=
$$- \frac{\pi}{4} - \frac{1}{10}$$
substitute to the expression
$$\cot^{2}{\left(x \right)} + \cot{\left(x \right)} \geq 0$$
$$\cot{\left(- \frac{\pi}{4} - \frac{1}{10} \right)} + \cot^{2}{\left(- \frac{\pi}{4} - \frac{1}{10} \right)} \geq 0$$
2/1 pi\ /1 pi\
cot |-- + --| - cot|-- + --| >= 0
\10 4 / \10 4 /
but
2/1 pi\ /1 pi\
cot |-- + --| - cot|-- + --| < 0
\10 4 / \10 4 /
Then
$$x \leq - \frac{\pi}{4}$$
no execute
one of the solutions of our inequality is:
$$x \geq - \frac{\pi}{4} \wedge x \leq \frac{\pi}{2}$$
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