Finding solutions to Trig Equations

[ashleem36]

Hi,

This is my first post and I'm looking for some help with a Trigonometry problem.

Find the solutions of tan²(2x)-tanx=3 over [0,2pi) to two decimal places.

I began by substituting the tan(2x) double angle identity (2tanx/1-tan²), but then things got really nasty after I performed the square and moved -tanx to the right side to cross multiply. I also tried graphing the equation and tracing the zeros, but this resulted in a large number of answers and I feel was not correct. Any help would be appreciated.

Ashlee

 

[DrPhil]

Hi,

This is my first post and I'm looking for some help with a Trigonometry problem.

Find the solutions of tan²(2x)-tanx=3 over [0,2pi) to two decimal places.

I began by substituting the tan(2x) double angle identity (2tanx/1-tan²), but then things got really nasty after I performed the square and moved -tanx to the right side to cross multiply. I also tried graphing the equation and tracing the zeros, but this resulted in a large number of answers and I feel was not correct. Any help would be appreciated.

Ashlee

Before I try to duplicate the nasty algebra you encountered, would you PLEASE check to be sure you have the problem copied correctly? In particular, the term \(\displaystyle \tan^2(2x)\) would be MUCH nicer if it were either \(\displaystyle \tan(2x)\) OR \(\displaystyle \tan^2x\), but not both!

 

[ashleem36]

Dr. Phil,

Unfortunately it is correct . It would be a much simpler problem if it were just tan(2x).

 

[DrPhil]

Dr. Phil,

Unfortunately it is correct . It would be a much simpler problem if it were just tan(2x).

Since you are to find the answer "to 2 decimal places," perhaps you could use numerical methods? Plotting just on the domain [0,2pi) should show hou how many roots there are, and approximately where. Do you know how to use calculus to approach the roots, or how to use successive approximations?

EDIT: I find six roots, three between [0,pi] and three more that are greater by pi.

 

[ashleem36]

No, I'm not familiar with either. I just sent the professor an email asking her to double check the equation (it was off a print out she gave us as a take home question) and make sure it was what she wanted to ask. I really feel there might have been a typo on her part because this does not resemble any questions we've gone over. Thank you for your time and I will post again with an update when I hear from her.

 

[ashleem36]

The equation is correct. The answer is due this morning and I still don't have it

 

[stapel]

The equation is correct.

Ick.

Okay; well, thanks for checking. Now, let's see what we can do....

My "messy algebra" ends up with this:

. . . . .\(\displaystyle \tan^5(x)\, -\, 3\tan^4(x)\, -\, 2\tan^3(x)\, -\, 10\tan^2(x)\, +\, \tan(x)\, +\, 3\, =\, 0\)

Assuming that the above is correct, you can check your decimal approximations (from your graphing calculator) here.

I have no idea how your instructor is expecting you to do this!

 

[ashleem36]

That's exactly what I got when I worked it out. I feel like she is going to show some kind of trick that would have made it an easier problem. She gives us the hardest questions. THANK YOU SO MUCH. Did you try graphing it on the calculator and finding the roots?

 

[Deleted member 4993]

May be your instructor wants you to get acquainted with wolframalpha.com...

Wolfram says there are three real and two complex roots.

 

[ashleem36]

May be your instructor wants you to get acquainted with wolframalpha.com...

Wolfram says there are three real and two complex roots.

Do I need to take arctan of these roots, or has wolfram already taken that into account? I'm definitely not familiar with wolfram.

 

[Deleted member 4993]

Do I need to take arctan of these roots, or has wolfram already taken that into account? I'm definitely not familiar with wolfram.

That will depend on "what" you posed as variable in the equation.

Did you look at the roots given by the software? The values are sort of self explanatory.

 

[DrPhil]

That will depend on "what" you posed as variable in the equation.

Did you look at the roots given by the software? The values are sort of self explanatory.

I gave Wolfram the original equation: \(\displaystyle \tan^2(2x) - \tan x = 3\), x = 0 to 2pi, and it gave me six real roots.

0.54, 1.00, 2.64, 3.68, 4.15, 5.78

The second three are the first ones + \(\displaystyle \pi\)

EDIT: the plot of the 5th order polynomial doesn't look very much like the original equation

 

[ashleem36]

Thank you for the help everyone. I spoke with other students in the class and everyone struggled with the question, and most just left it blank. At the end of class I asked the professor to work out the problem for me and she ended with:

tan^3x-3tan^2x-5tanx+3=0

u^3-3u^2-5u+3=0 where u=tanx

She then graphed this equation to find the roots, one of which was tanx=.482. Then take the arctan(.482) as one solution and then add pi. Then follow same procedure for other roots. She was working on it and was surprised that she didn't give us an equation that factored nicely :???:

I don't think anyone got the answer, so I'm not sure what she'll do. Again, thank you for the help.

 

[Deleted member 4993]

I gave Wolfram the original equation: \(\displaystyle \tan^2(2x) - \tan x = 3\), x = 0 to 2pi, and it gave me six real roots.

0.54, 1.00, 2.64, 3.68, 4.15, 5.78

The second three are the first ones + \(\displaystyle \pi\)

EDIT: the plot of the 5th order polynomial doesn't look very much like the original equation

Those roots are then for 'x' - no need to take arctan.