Mister Exam
Other calculators
- Square Inequality Step-by-Step
- Inequality with the module Step by Step
- Logarithmic inequality online
-
How to use it?
- Inequation:
- y>/x+1
- 5^(1-x)>0,2
- logsqrt3x<=2
- (|x|)+(|x-3|)>=3
- Derivative of:
-
ctg2x
- Integral of d{x}:
-
ctg2x
-
Identical expressions
- ctg2x< one
- ctg2x less than 1
- ctg2x less than one
-
Similar expressions
- ctg(2x+pi)≥-sqrt(3)
- ctg(2x)≤√3
- ctg(2x)>=0
- arctg^2(x)-4arctgx+3>0
ctg2x<1 inequation
The solution
Detail solution
Given the inequality:
$$\cot{\left(2 x \right)} < 1$$
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:
$$x_{0} < x_{1}$$
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
$$\cot{\left(2 x \right)} < 1$$
$$\cot{\left(2 \left(- \frac{1}{10} + \frac{\pi}{8}\right) \right)} < 1$$
but
Then
$$x < \frac{\pi}{8}$$
no execute
the solution of our inequality is:
$$x > \frac{\pi}{8}$$
$$\cot{\left(2 x \right)} < 1$$
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:
$$x_{0} < x_{1}$$
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
$$\cot{\left(2 x \right)} < 1$$
$$\cot{\left(2 \left(- \frac{1}{10} + \frac{\pi}{8}\right) \right)} < 1$$
/1 pi\ tan|- + --| < 1 \5 4 /
but
/1 pi\ tan|- + --| > 1 \5 4 /
Then
$$x < \frac{\pi}{8}$$
no execute
the solution of our inequality is:
$$x > \frac{\pi}{8}$$
_____
/
-------ο-------
x1
Rapid solution
[src]
/ / ___________\ \ | | / ___ | | | pi |\/ 2 - \/ 2 | | And|x < --, atan|--------------| < x| | 2 | ___________| | | | / ___ | | \ \\/ 2 + \/ 2 / /
$$x < \frac{\pi}{2} \wedge \operatorname{atan}{\left(\frac{\sqrt{2 - \sqrt{2}}}{\sqrt{\sqrt{2} + 2}} \right)} < x$$
(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.open(atan(sqrt(2 - sqrt(2))/sqrt(sqrt(2) + 2)), pi/2)