ctg4x<=sqrt3/3 inequation

Given the inequality:
$$\cot{\left(4 x \right)} \leq \frac{\sqrt{3}}{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot{\left(4 x \right)} = \frac{\sqrt{3}}{3}$$
Solve:
Given the equation
$$\cot{\left(4 x \right)} = \frac{\sqrt{3}}{3}$$
transform
$$\cot{\left(4 x \right)} - 1 - \frac{\sqrt{3}}{3} = 0$$
$$\cot{\left(4 x \right)} - 1 - \frac{\sqrt{3}}{3} = 0$$
Do replacement
$$w = \cot{\left(4 x \right)}$$
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
$$\cot{\left(4 x \right)} = w$$
substitute w:
$$x_{1} = \frac{\pi}{12}$$
$$x_{1} = \frac{\pi}{12}$$
This roots
$$x_{1} = \frac{\pi}{12}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\pi}{12}$$
=
$$- \frac{1}{10} + \frac{\pi}{12}$$
substitute to the expression
$$\cot{\left(4 x \right)} \leq \frac{\sqrt{3}}{3}$$
$$\cot{\left(4 \left(- \frac{1}{10} + \frac{\pi}{12}\right) \right)} \leq \frac{\sqrt{3}}{3}$$

                 ___
   /2   pi\    \/ 3 
tan|- + --| <= -----
   \5   6 /      3  
               

but

                 ___
   /2   pi\    \/ 3 
tan|- + --| >= -----
   \5   6 /      3  
               

Then
$$x \leq \frac{\pi}{12}$$
no execute
the solution of our inequality is:
$$x \geq \frac{\pi}{12}$$

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