Mister Exam
Other calculators
- Square Inequality Step-by-Step
- Inequality with the module Step by Step
- Logarithmic inequality online
-
How to use it?
- Inequation:
- (1/(x-1))-(1/(x-2))=>(1/(x+1))-(1/(x+2))
- (x^2(x+2))/(2x+3)<=0
- 7x<-3
- (3/(x+1))<=(5/(x+2))
- Graphing y =:
-
ctg(2x)
- Integral of d{x}:
- ctg(2x)
- Derivative of:
-
ctg(2x)
-
Identical expressions
- ctg(2x)<= one
- ctg(2x) less than or equal to 1
- ctg(2x) less than or equal to one
- ctg2x<=1
-
Similar expressions
- ctg((2x/3)-(п/5))≥-1
- (ctg(2*x))^2-3ctg(x)+2>=0
- arctg^2(x)/n^3<=1/n^3
ctg(2x)<=1 inequation
The solution
Detail solution
Given the inequality:
$$\cot{\left(2 x \right)} \leq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot{\left(2 x \right)} = 1$$
Solve:
Given the equation
$$\cot{\left(2 x \right)} = 1$$
transform
$$\cot{\left(2 x \right)} - 1 = 0$$
$$\cot{\left(2 x \right)} - 1 = 0$$
Do replacement
$$w = \cot{\left(2 x \right)}$$
Move free summands (without w)
from left part to right part, we given:
$$w = 1$$
We get the answer: w = 1
do backward replacement
$$\cot{\left(2 x \right)} = w$$
substitute w:
$$x_{1} = \frac{\pi}{8}$$
$$x_{1} = \frac{\pi}{8}$$
This roots
$$x_{1} = \frac{\pi}{8}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\pi}{8}$$
=
$$- \frac{1}{10} + \frac{\pi}{8}$$
substitute to the expression
$$\cot{\left(2 x \right)} \leq 1$$
$$\cot{\left(2 \left(- \frac{1}{10} + \frac{\pi}{8}\right) \right)} \leq 1$$
but
Then
$$x \leq \frac{\pi}{8}$$
no execute
the solution of our inequality is:
$$x \geq \frac{\pi}{8}$$
$$\cot{\left(2 x \right)} \leq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot{\left(2 x \right)} = 1$$
Solve:
Given the equation
$$\cot{\left(2 x \right)} = 1$$
transform
$$\cot{\left(2 x \right)} - 1 = 0$$
$$\cot{\left(2 x \right)} - 1 = 0$$
Do replacement
$$w = \cot{\left(2 x \right)}$$
Move free summands (without w)
from left part to right part, we given:
$$w = 1$$
We get the answer: w = 1
do backward replacement
$$\cot{\left(2 x \right)} = w$$
substitute w:
$$x_{1} = \frac{\pi}{8}$$
$$x_{1} = \frac{\pi}{8}$$
This roots
$$x_{1} = \frac{\pi}{8}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\pi}{8}$$
=
$$- \frac{1}{10} + \frac{\pi}{8}$$
substitute to the expression
$$\cot{\left(2 x \right)} \leq 1$$
$$\cot{\left(2 \left(- \frac{1}{10} + \frac{\pi}{8}\right) \right)} \leq 1$$
/1 pi\ tan|- + --| <= 1 \5 4 /
but
/1 pi\ tan|- + --| >= 1 \5 4 /
Then
$$x \leq \frac{\pi}{8}$$
no execute
the solution of our inequality is:
$$x \geq \frac{\pi}{8}$$
_____
/
-------•-------
x1
Rapid solution
[src]
/ / ___________\ \ | | / ___ | | | |\/ 2 - \/ 2 | pi| And|atan|--------------| <= x, x < --| | | ___________| 2 | | | / ___ | | \ \\/ 2 + \/ 2 / /
$$\operatorname{atan}{\left(\frac{\sqrt{2 - \sqrt{2}}}{\sqrt{\sqrt{2} + 2}} \right)} \leq x \wedge x < \frac{\pi}{2}$$
(x < pi/2)∧(atan(sqrt(2 - sqrt(2))/sqrt(2 + sqrt(2))) <= x)
Rapid solution 2
[src]
/ ___________\
| / ___ |
|\/ 2 - \/ 2 | pi
[atan|--------------|, --)
| ___________| 2
| / ___ |
\\/ 2 + \/ 2 /
$$x\ in\ \left[\operatorname{atan}{\left(\frac{\sqrt{2 - \sqrt{2}}}{\sqrt{\sqrt{2} + 2}} \right)}, \frac{\pi}{2}\right)$$
x in Interval.Ropen(atan(sqrt(2 - sqrt(2))/sqrt(sqrt(2) + 2)), pi/2)