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ctgx<=-√3 inequation

A inequation with variable

The solution

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cot(x) <= -\/ 3 
cot(x)3\cot{\left(x \right)} \leq - \sqrt{3}
cot(x) <= -sqrt(3)
Detail solution
Given the inequality:
cot(x)3\cot{\left(x \right)} \leq - \sqrt{3}
To solve this inequality, we must first solve the corresponding equation:
cot(x)=3\cot{\left(x \right)} = - \sqrt{3}
Solve:
Given the equation
cot(x)=3\cot{\left(x \right)} = - \sqrt{3}
transform
cot(x)1+3=0\cot{\left(x \right)} - 1 + \sqrt{3} = 0
cot(x)1+3=0\cot{\left(x \right)} - 1 + \sqrt{3} = 0
Do replacement
w=cot(x)w = \cot{\left(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+3=1w + \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(x)=w\cot{\left(x \right)} = w
substitute w:
x1=π6x_{1} = - \frac{\pi}{6}
x1=π6x_{1} = - \frac{\pi}{6}
This roots
x1=π6x_{1} = - \frac{\pi}{6}
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
x0x1x_{0} \leq x_{1}
For example, let's take the point
x0=x1110x_{0} = x_{1} - \frac{1}{10}
=
π6110- \frac{\pi}{6} - \frac{1}{10}
=
π6110- \frac{\pi}{6} - \frac{1}{10}
substitute to the expression
cot(x)3\cot{\left(x \right)} \leq - \sqrt{3}
cot(π6110)3\cot{\left(- \frac{\pi}{6} - \frac{1}{10} \right)} \leq - \sqrt{3}
    /1    pi\       ___
-cot|-- + --| <= -\/ 3 
    \10   6 /    

but
    /1    pi\       ___
-cot|-- + --| >= -\/ 3 
    \10   6 /    

Then
xπ6x \leq - \frac{\pi}{6}
no execute
the solution of our inequality is:
xπ6x \geq - \frac{\pi}{6}
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        /
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       x1
Rapid solution 2 [src]
 5*pi     
[----, pi)
  6       
x in [5π6,π)x\ in\ \left[\frac{5 \pi}{6}, \pi\right)
x in Interval.Ropen(5*pi/6, pi)
Rapid solution [src]
   /5*pi             \
And|---- <= x, x < pi|
   \ 6               /
5π6xx<π\frac{5 \pi}{6} \leq x \wedge x < \pi
(x < pi)∧(5*pi/6 <= x)