ctg(5x)>sqrt(3)/3 inequation

Given the inequality:
$$\cot{\left(5 x \right)} > \frac{\sqrt{3}}{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot{\left(5 x \right)} = \frac{\sqrt{3}}{3}$$
Solve:
Given the equation
$$\cot{\left(5 x \right)} = \frac{\sqrt{3}}{3}$$
transform
$$\cot{\left(5 x \right)} - 1 - \frac{\sqrt{3}}{3} = 0$$
$$\cot{\left(5 x \right)} - 1 - \frac{\sqrt{3}}{3} = 0$$
Do replacement
$$w = \cot{\left(5 x \right)}$$
Expand brackets in the left part

-1 + w - sqrt3/3 = 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
$$\cot{\left(5 x \right)} = w$$
substitute w:
$$x_{1} = \frac{\pi}{15}$$
$$x_{1} = \frac{\pi}{15}$$
This roots
$$x_{1} = \frac{\pi}{15}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\pi}{15}$$
=
$$- \frac{1}{10} + \frac{\pi}{15}$$
substitute to the expression
$$\cot{\left(5 x \right)} > \frac{\sqrt{3}}{3}$$
$$\cot{\left(5 \left(- \frac{1}{10} + \frac{\pi}{15}\right) \right)} > \frac{\sqrt{3}}{3}$$

                ___
   /1   pi\   \/ 3 
tan|- + --| > -----
   \2   6 /     3  
              

the solution of our inequality is:
$$x < \frac{\pi}{15}$$

 _____          
      \    
-------ο-------
       x1