sqrt(3)ctg(x)+3>=2 inequation

Given the inequality:
$$\sqrt{3} \cot{\left(x \right)} + 3 \geq 2$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{3} \cot{\left(x \right)} + 3 = 2$$
Solve:
Given the equation
$$\sqrt{3} \cot{\left(x \right)} + 3 = 2$$
transform
$$\sqrt{3} \cot{\left(x \right)} + 1 = 0$$
$$\left(\sqrt{3} \cot{\left(x \right)} + 3\right) - 2 = 0$$
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:
$$\sqrt{3} w = -1$$
Divide both parts of the equation by sqrt(3)

w = -1 / (sqrt(3))

We get the answer: w = -sqrt(3)/3
do backward replacement
$$\cot{\left(x \right)} = w$$
substitute w:
$$x_{1} = - \frac{\pi}{3}$$
$$x_{1} = - \frac{\pi}{3}$$
This roots
$$x_{1} = - \frac{\pi}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{3} - \frac{1}{10}$$
=
$$- \frac{\pi}{3} - \frac{1}{10}$$
substitute to the expression
$$\sqrt{3} \cot{\left(x \right)} + 3 \geq 2$$
$$\sqrt{3} \cot{\left(- \frac{\pi}{3} - \frac{1}{10} \right)} + 3 \geq 2$$

      ___    /1    pi\     
3 - \/ 3 *cot|-- + --| >= 2
             \10   3 /     

the solution of our inequality is:
$$x \leq - \frac{\pi}{3}$$

 _____          
      \    
-------•-------
       x1