Mister Exam

Lang:

How to use it?
Inequation:
√(x^2-2x)>4-x
logx+1,(x-2)<=1
|x|+|y|<3
log(1/3)x≤−1.
Derivative of:
cot(x)
Integral of d{x}:
cot(x)
Limit of the function:
cot(x)
Identical expressions
cot(x)>(sqrt(three))* one / three
cotangent of (x) greater than ( square root of (3)) multiply by 1 divide by 3
cotangent of (x) greater than ( square root of (three)) multiply by one divide by three
cot(x)>(√(3))*1/3
cot(x)>(sqrt(3))1/3
cotx>sqrt31/3
cot(x)>(sqrt(3))*1 divide by 3
Similar expressions
tg(1/2x)=<-sqrt(3)*1/3
cot(x/2)>=1/sqrt(3)
cot(x)>=(-sqrt(3))*1/3
cot(x/2)>=sqrt(3)
cot(x)<=-sqrt(3)*1/3
cotx>=1

cot(x)>(sqrt(3))*1/3 inequation

You have entered [src]

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         \/ 3 
cot(x) > -----
           3

\cot{\left(x \right)} > \frac{\sqrt{3}}{3}

cot(x) > sqrt(3)/3

Detail solution

Given the inequality:

\cot{\left(x \right)} > \frac{\sqrt{3}}{3}

To solve this inequality, we must first solve the corresponding equation:

\cot{\left(x \right)} = \frac{\sqrt{3}}{3}

Solve:
Given the equation

\cot{\left(x \right)} = \frac{\sqrt{3}}{3}

transform

\cot{\left(x \right)} - 1 - \frac{\sqrt{3}}{3} = 0

\cot{\left(x \right)} - 1 - \frac{\sqrt{3}}{3} = 0

Do replacement

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 - \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(x \right)} = w

substitute w:

x_{1} = \frac{\pi}{3}

x_{1} = \frac{\pi}{3}

This roots

x_{1} = \frac{\pi}{3}

is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:

x_{0} < x_{1}

For example, let's take the point

x_{0} = x_{1} - \frac{1}{10}

- \frac{1}{10} + \frac{\pi}{3}

- \frac{1}{10} + \frac{\pi}{3}

substitute to the expression

\cot{\left(x \right)} > \frac{\sqrt{3}}{3}

\cot{\left(- \frac{1}{10} + \frac{\pi}{3} \right)} > \frac{\sqrt{3}}{3}

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

the solution of our inequality is:

x < \frac{\pi}{3}

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       x1

Rapid solution 2 [src]

    pi 
(0, --)
    3

x\ in\ \left(0, \frac{\pi}{3}\right)

x in Interval.open(0, pi/3)

Rapid solution [src]

   /           pi\
And|0 < x, x < --|
   \           3 /

0 < x \wedge x < \frac{\pi}{3}

(0 < x)∧(x < pi/3)