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How to use it?
- Inequation:
-
tan(x+pi/3)<1
- 1+5/(log4(x)-3)+6/((log4(x))^2-log4(64*x^6)+12)>0
- (2x^2-5x+2)<=0
- (x+8)(3-x)(1,5-x)<0
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Identical expressions
- tan(x+pi/ three)< one
- tangent of (x plus Pi divide by 3) less than 1
- tangent of (x plus Pi divide by three) less than one
- tanx+pi/3<1
- tan(x+pi divide by 3)<1
-
Similar expressions
- tan(x-pi/3)<1
- tan((x+pi)/3)>0
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What you mean?
- tan((x + pi)/3) < 1
- tan(x + pi/3) < 1
- tan(x + pi/3) < 1
tan(x+pi/3)<1 inequation
The solution
You have entered
[src]
/ pi\ tan|x + --| < 1 \ 3 /
$$\tan{\left(x + \frac{\pi}{3} \right)} < 1$$
tan(x + pi/3) < 1
Detail solution
Given the inequality:
$$\tan{\left(x + \frac{\pi}{3} \right)} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(x + \frac{\pi}{3} \right)} = 1$$
Solve:
Given the equation
$$\tan{\left(x + \frac{\pi}{3} \right)} = 1$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x + \frac{\pi}{3} = \pi n + \operatorname{atan}{\left(1 \right)}$$
Or
$$x + \frac{\pi}{3} = \pi n + \frac{\pi}{4}$$
, where n - is a integer
Move
$$\frac{\pi}{3}$$
to right part of the equation with the opposite sign, in total:
$$x = \pi n - \frac{\pi}{12}$$
$$x_{1} = \pi n - \frac{\pi}{12}$$
$$x_{1} = \pi n - \frac{\pi}{12}$$
This roots
$$x_{1} = \pi n - \frac{\pi}{12}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n - \frac{\pi}{12}\right) - \frac{1}{10}$$
=
$$\pi n - \frac{\pi}{12} - \frac{1}{10}$$
substitute to the expression
$$\tan{\left(x + \frac{\pi}{3} \right)} < 1$$
$$\tan{\left(\left(\pi n - \frac{\pi}{12} - \frac{1}{10}\right) + \frac{\pi}{3} \right)} < 1$$
the solution of our inequality is:
$$x < \pi n - \frac{\pi}{12}$$
$$\tan{\left(x + \frac{\pi}{3} \right)} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(x + \frac{\pi}{3} \right)} = 1$$
Solve:
Given the equation
$$\tan{\left(x + \frac{\pi}{3} \right)} = 1$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x + \frac{\pi}{3} = \pi n + \operatorname{atan}{\left(1 \right)}$$
Or
$$x + \frac{\pi}{3} = \pi n + \frac{\pi}{4}$$
, where n - is a integer
Move
$$\frac{\pi}{3}$$
to right part of the equation with the opposite sign, in total:
$$x = \pi n - \frac{\pi}{12}$$
$$x_{1} = \pi n - \frac{\pi}{12}$$
$$x_{1} = \pi n - \frac{\pi}{12}$$
This roots
$$x_{1} = \pi n - \frac{\pi}{12}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n - \frac{\pi}{12}\right) - \frac{1}{10}$$
=
$$\pi n - \frac{\pi}{12} - \frac{1}{10}$$
substitute to the expression
$$\tan{\left(x + \frac{\pi}{3} \right)} < 1$$
$$\tan{\left(\left(\pi n - \frac{\pi}{12} - \frac{1}{10}\right) + \frac{\pi}{3} \right)} < 1$$
/1 pi\ cot|-- + --| < 1 \10 4 /
the solution of our inequality is:
$$x < \pi n - \frac{\pi}{12}$$
_____
\
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x_1
Solving inequality on a graph
Rapid solution
[src]
/ / ___ ___ \\ |pi | \/ 6 - \/ 2 || And|-- < x, x < pi + atan|---------------|| |6 | ___ ___|| \ \- \/ 2 - \/ 6 //
$$\frac{\pi}{6} < x \wedge x < \operatorname{atan}{\left(\frac{- \sqrt{2} + \sqrt{6}}{- \sqrt{6} - \sqrt{2}} \right)} + \pi$$
(pi/6 < x)∧(x < pi + atan((sqrt(6) - sqrt(2))/(-sqrt(2) - sqrt(6))))
Rapid solution 2
[src]
/ ___ ___\
pi |\/ 2 - \/ 6 |
(--, pi + atan|-------------|)
6 | ___ ___|
\\/ 2 + \/ 6 /
$$x\ in\ \left(\frac{\pi}{6}, \operatorname{atan}{\left(\frac{- \sqrt{6} + \sqrt{2}}{\sqrt{2} + \sqrt{6}} \right)} + \pi\right)$$
x in Interval.open(pi/6, atan((-sqrt(6) + sqrt(2))/(sqrt(2) + sqrt(6))) + pi)
The graph