ctg(x)≥√3 inequation

Given the inequality:
$$\cot{\left(x \right)} \geq \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot{\left(x \right)} = \sqrt{3}$$
Solve:
Given the equation
$$\cot{\left(x \right)} = \sqrt{3}$$
transform
$$\cot{\left(x \right)} - \sqrt{3} - 1 = 0$$
$$\cot{\left(x \right)} - \sqrt{3} - 1 = 0$$
Do replacement
$$w = \cot{\left(x \right)}$$
Expand brackets in the left part

-1 + w - sqrt3 = 0

Move free summands (without w)
from left part to right part, we given:
$$w - \sqrt{3} = 1$$
Divide both parts of the equation by (w - sqrt(3))/w

w = 1 / ((w - sqrt(3))/w)

We get the answer: w = 1 + sqrt(3)
do backward replacement
$$\cot{\left(x \right)} = w$$
substitute w:
$$x_{1} = \frac{\pi}{6}$$
$$x_{1} = \frac{\pi}{6}$$
This roots
$$x_{1} = \frac{\pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\pi}{6}$$
=
$$- \frac{1}{10} + \frac{\pi}{6}$$
substitute to the expression
$$\cot{\left(x \right)} \geq \sqrt{3}$$
$$\cot{\left(- \frac{1}{10} + \frac{\pi}{6} \right)} \geq \sqrt{3}$$

   /1    pi\      ___
tan|-- + --| >= \/ 3 
   \10   3 /    

the solution of our inequality is:
$$x \leq \frac{\pi}{6}$$

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