Mister Exam
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How to use it?
- Inequation:
- x^2>-x-2
- (x+-3)/x+1(2x+4)_>0
- sqrt(5^x-625)+log^2x+x<=6
-
arctg(x)>=1
-
Identical expressions
- tg(t)>sqrt three /3
- tg(t) greater than square root of 3 divide by 3
- tg(t) greater than square root of three divide by 3
- tg(t)>√3/3
- tgt>sqrt3/3
- tg(t)>sqrt3 divide by 3
-
Similar expressions
- cot(x+pi/6)>=sqrt(3)/3
- tgt>√3/3
- tg(2x-pi/3)<sqrt(3)/3
- cot((pi/4)-2*x)>sqrt(3)/3
- tgt<0
- tgt>sqrt3/3
tg(t)>sqrt3/3 inequation
The solution
You have entered
[src]
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\/ 3
tan(t) > -----
3
$$\tan{\left(t \right)} > \frac{\sqrt{3}}{3}$$
tan(t) > sqrt(3)/3
Detail solution
Given the inequality:
$$\tan{\left(t \right)} > \frac{\sqrt{3}}{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(t \right)} = \frac{\sqrt{3}}{3}$$
Solve:
Given the equation
$$\tan{\left(t \right)} = \frac{\sqrt{3}}{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$t = \pi n + \operatorname{atan}{\left(\frac{\sqrt{3}}{3} \right)}$$
Or
$$t = \pi n + \frac{\pi}{6}$$
, where n - is a integer
$$t_{1} = \pi n + \frac{\pi}{6}$$
$$t_{1} = \pi n + \frac{\pi}{6}$$
This roots
$$t_{1} = \pi n + \frac{\pi}{6}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$t_{0} < t_{1}$$
For example, let's take the point
$$t_{0} = t_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{6}$$
substitute to the expression
$$\tan{\left(t \right)} > \frac{\sqrt{3}}{3}$$
$$\tan{\left(\pi n - \frac{1}{10} + \frac{\pi}{6} \right)} > \frac{\sqrt{3}}{3}$$
Then
$$t < \pi n + \frac{\pi}{6}$$
no execute
the solution of our inequality is:
$$t > \pi n + \frac{\pi}{6}$$
$$\tan{\left(t \right)} > \frac{\sqrt{3}}{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(t \right)} = \frac{\sqrt{3}}{3}$$
Solve:
Given the equation
$$\tan{\left(t \right)} = \frac{\sqrt{3}}{3}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$t = \pi n + \operatorname{atan}{\left(\frac{\sqrt{3}}{3} \right)}$$
Or
$$t = \pi n + \frac{\pi}{6}$$
, where n - is a integer
$$t_{1} = \pi n + \frac{\pi}{6}$$
$$t_{1} = \pi n + \frac{\pi}{6}$$
This roots
$$t_{1} = \pi n + \frac{\pi}{6}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$t_{0} < t_{1}$$
For example, let's take the point
$$t_{0} = t_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \frac{\pi}{6}$$
substitute to the expression
$$\tan{\left(t \right)} > \frac{\sqrt{3}}{3}$$
$$\tan{\left(\pi n - \frac{1}{10} + \frac{\pi}{6} \right)} > \frac{\sqrt{3}}{3}$$
___
/ 1 pi \ \/ 3
tan|- -- + -- + pi*n| > -----
\ 10 6 / 3
Then
$$t < \pi n + \frac{\pi}{6}$$
no execute
the solution of our inequality is:
$$t > \pi n + \frac{\pi}{6}$$
_____
/
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t1
Solving inequality on a graph
Rapid solution
[src]
/pi pi\ And|-- < t, t < --| \6 2 /
$$\frac{\pi}{6} < t \wedge t < \frac{\pi}{2}$$
(pi/6 < t)∧(t < pi/2)
Rapid solution 2
[src]
pi pi (--, --) 6 2
$$t\ in\ \left(\frac{\pi}{6}, \frac{\pi}{2}\right)$$
t in Interval.open(pi/6, pi/2)