ctg(2x)≤√3 inequation

Given the inequality:
$$\cot{\left(2 x \right)} \leq \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot{\left(2 x \right)} = \sqrt{3}$$
Solve:
Given the equation
$$\cot{\left(2 x \right)} = \sqrt{3}$$
transform
$$\cot{\left(2 x \right)} - \sqrt{3} - 1 = 0$$
$$\cot{\left(2 x \right)} - \sqrt{3} - 1 = 0$$
Do replacement
$$w = \cot{\left(2 x \right)}$$
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
$$\cot{\left(2 x \right)} = w$$
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
$$\cot{\left(2 x \right)} \leq \sqrt{3}$$
$$\cot{\left(2 \left(- \frac{1}{10} + \frac{\pi}{12}\right) \right)} \leq \sqrt{3}$$

   /1   pi\      ___
tan|- + --| <= \/ 3 
   \5   3 /    

but

   /1   pi\      ___
tan|- + --| >= \/ 3 
   \5   3 /    

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

         _____  
        /
-------•-------
       x1