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Solving inequalities/ ctgx<√3/3

ctgx<√3/3 inequation

A inequation with variable

The solution

You have entered [src] 
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         \/ 3 
cot(x) < -----
           3
cot⁡(x)<33\cot{\left(x \right)} < \frac{\sqrt{3}}{3} 
cot(x) < sqrt(3)/3
Detail solution
Given the inequality:
cot⁡(x)<33\cot{\left(x \right)} < \frac{\sqrt{3}}{3}
To solve this inequality, we must first solve the corresponding equation:
cot⁡(x)=33\cot{\left(x \right)} = \frac{\sqrt{3}}{3}
Solve:
Given the equation
cot⁡(x)=33\cot{\left(x \right)} = \frac{\sqrt{3}}{3}
transform
cot⁡(x)−1−33=0\cot{\left(x \right)} - 1 - \frac{\sqrt{3}}{3} = 0
cot⁡(x)−1−33=0\cot{\left(x \right)} - 1 - \frac{\sqrt{3}}{3} = 0
Do replacement
w=cot⁡(x)w = \cot{\left(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−33=1w - \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⁡(x)=w\cot{\left(x \right)} = w
substitute w:
x1=π3x_{1} = \frac{\pi}{3}
x1=π3x_{1} = \frac{\pi}{3}
This roots
x1=π3x_{1} = \frac{\pi}{3}
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
x0<x1x_{0} < x_{1}
For example, let's take the point
x0=x1−110x_{0} = x_{1} - \frac{1}{10}
=
−110+π3- \frac{1}{10} + \frac{\pi}{3}
=
−110+π3- \frac{1}{10} + \frac{\pi}{3}
substitute to the expression
cot⁡(x)<33\cot{\left(x \right)} < \frac{\sqrt{3}}{3}
cot⁡(−110+π3)<33\cot{\left(- \frac{1}{10} + \frac{\pi}{3} \right)} < \frac{\sqrt{3}}{3}
                 ___
   /1    pi\   \/ 3 
tan|-- + --| < -----
   \10   6 /     3

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

Then
x<π3x < \frac{\pi}{3}
no execute
the solution of our inequality is:
x>π3x > \frac{\pi}{3}
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       x1
Rapid solution [src] 
   /pi            \
And|-- < x, x < pi|
   \3             /
π3<x∧x<π\frac{\pi}{3} < x \wedge x < \pi 
(x < pi)∧(pi/3 < x)
Rapid solution 2 [src] 
 pi     
(--, pi)
 3
x in (π3,π)x\ in\ \left(\frac{\pi}{3}, \pi\right) 
x in Interval.open(pi/3, pi)
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