cot(x)/3>=1 inequation

Given the inequality:
$$\frac{\cot{\left(x \right)}}{3} \geq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\cot{\left(x \right)}}{3} = 1$$
Solve:
Given the equation
$$\frac{\cot{\left(x \right)}}{3} = 1$$
transform
$$\frac{\cot{\left(x \right)}}{3} - 1 = 0$$
$$\frac{\cot{\left(x \right)}}{3} - 1 = 0$$
Do replacement
$$w = \cot{\left(x \right)}$$
Move free summands (without w)
from left part to right part, we given:
$$\frac{w}{3} = 1$$
Divide both parts of the equation by 1/3

w = 1 / (1/3)

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

-cot(1/10 - acot(3))      
--------------------- >= 1
          3               

the solution of our inequality is:
$$x \leq \operatorname{acot}{\left(3 \right)}$$

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