Solve Trig Equation #2 on 0<=theta<=2pi

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Homework Help Overview

The discussion revolves around solving the trigonometric equation tan²x - 4tanx = 0 within the interval 0 ≤ θ ≤ 2π. Participants explore various methods for finding the solutions, including factoring and graphical approaches.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss factoring the equation and consider the implications of tanx = 4 and tanx = 0. Some suggest using graphical methods to identify solutions, while others express uncertainty about the methods learned in class.

Discussion Status

There is an ongoing exploration of different approaches to the problem, with some participants providing potential solutions and others questioning the methods due to their current understanding. Guidance is offered regarding the periodic nature of the tangent function and the relationship between sine and tangent.

Contextual Notes

Some participants indicate that they have not yet learned certain concepts, such as arctan, which affects their ability to fully engage with the problem. There is a mention of needing to find solutions within specific constraints of the interval.

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Homework Statement


Solve for x on the interval 0<= theta <= 2pi. Give exact solutions if possible. Otherwise round to the nearest hundredth of a radian.

Homework Equations


tan^2x-4tanx = 0

The Attempt at a Solution



Tried factoring it to (tanx-4)(tanx) = 0

tanx=4 or 0 I guess? Not sure where to go from there

or instead of factoring I could immediately make it sin/cos, which gives

(sin^2x - 4sincos) / (cos^2x) = 0
 
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I would plot a graph of tan(x) for x \in [0,\pi]...at what points does tanx=4?...At what points does tanx=0?...Those will be your solutions.
 
"Tried factoring it to (tanx-4)(tanx) = 0"

@Draggu now it is good. So as you said there are two solutions: tanx=4 and tanx=0.

The solutions for x are:

x1=[arctan(4)+kπ]

x2=[arctan(0)+kπ]

Regards.
 
Дьявол said:
"Tried factoring it to (tanx-4)(tanx) = 0"

@Draggu now it is good. So as you said there are two solutions: tanx=4 and tanx=0.

The solutions for x are:

x1=[arctan(4)+kπ]

x2=[arctan(0)+kπ]

Regards.

We have not learned that yet in class... so doing that wouldn't work. :p
 
gabbagabbahey said:
I would plot a graph of tan(x) for x \in [0,\pi]...at what points does tanx=4?...At what points does tanx=0?...Those will be your solutions.

There is no exact radian measure/degree where tanx=4
Tanx=4 approx at 76degrees. God I'm lost.
 
Well I got: (19pi/45), (64pi/45), pi, 2pi as the answers
 
Draggu said:
We have not learned that yet in class... so doing that wouldn't work. :p
You know how to draw graphic of tan function, and do not know how to find arctan(x)?

That's strange.

Anyway, you got the solutions correctly, since the task is to find the solutions in the interval [0,2п]. In the future, you will learn about periods.So if you want to be more correctly :smile: you should write 19п/45 + kп and п + kп, where k Є Z
 
Well, for tan(x) = 0, you don't need to know tan(x). tan(x) = \frac{sin(x)}{cos(x)} so tan(x) = 0 when sin(x) = 0.
 

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