ctgx>-(1/√3) inequation

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

-1 + w + 1/sqrt+1/3) = 0

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

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

We get the answer: w = 1 - 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:
$$x_{0} < x_{1}$$
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
$$\cot{\left(x \right)} > - \frac{1}{\sqrt{3}}$$
$$\cot{\left(- \frac{\pi}{3} - \frac{1}{10} \right)} > - \frac{1}{\sqrt{3}}$$

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

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

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