Trigonometric Equation

Printable View

August 18th 2011, 04:54 PM
dagbayani481

Trigonometric Equation

So the questions asks find all the values of x in interval [0,2pi)

for the following equation $8 + 4 cos(2x) = 12 cos(x)$

So the way I go about solving it is I subtract 12cos(x) to the other side and replace cos(2x) for the double angle identity which gives me:

$4 (2cos^2x-1) -12 cosx +8=0$

So essentially, I distribute the 4 and rewrite the equation as follows:

$8cos^2x -12 cosx +4=0$

Next I take out the common factor of 4 and factor out the whole equation:

$4 (2cos^2x -3 cosx + 1)=0$ to $(2cosx-1)(cosx-1)=0$

After solving each one I get the following answers: $x = \frac {\pi}{3}, \frac {5\pi}{3}, & 0$

However, the online assignment guide is telling me I've gotten the wrong answers. I'm honestly lost, can anyone tell me where exactly I'm getting this wrong?
August 18th 2011, 07:34 PM
Quacky

Re: Trigonometric Equation

Quote:

Originally Posted by dagbayani481

So the questions asks find all the values of x in interval [0,2pi)

for the following equation $8 + 4 cos(2x) = 12 cos(x)$

So the way I go about solving it is I subtract 12cos(x) to the other side and replace cos(2x) for the double angle identity which gives me:

$4 (2cos^2x-1) -12 cosx +8=0$

So essentially, I distribute the 4 and rewrite the equation as follows:

$8cos^2x -12 cosx +4=0$

Next I take out the common factor of 4 and factor out the whole equation:

$4 (2cos^2x -3 cosx + 1)=0$ to $(2cosx-1)(cosx-1)=0$

After solving each one I get the following answers: $x = \frac {\pi}{3}, \frac {5\pi}{3}, & 0$

However, the online assignment guide is telling me I've gotten the wrong answers. I'm honestly lost, can anyone tell me where exactly I'm getting this wrong?

I agree with your answers, but I honestly struggle to believe that they would give you a tough trigonometric equation with a common factor at the start - why have they just randomly taken something and multiplied it by 4? I suspect it could be an error in the question. What answers are you supposed to get? Is this the full question?
August 18th 2011, 08:40 PM
dagbayani481

Re: Trigonometric Equation

Quote:

Originally Posted by Quacky

I agree with your answers, but I honestly struggle to believe that they would give you a tough trigonometric equation with a common factor at the start - why have they just randomly taken something and multiplied it by 4? I suspect it could be an error in the question. What answers are you supposed to get? Is this the full question?

I think the reasoning behind the equation is the fact that this is pre-calculus review in a calculus course. I'm assuming the difficulty of the problem is based on that fact. I'm starting to believe there might be an error in the question and I may bring it up to the instructor this Monday. Sadly, the problem does not provide an answer to the solution. In the meantime, I'm also posting a picture of the problem directly from the website to see if you guys can find any mistakes with my own solution:

http://img841.imageshack.us/img841/242/wronganswer.jpg
August 18th 2011, 08:52 PM
Prove It

Re: Trigonometric Equation

Surely $\displaystyle x = 2\pi$ is a solution as well :)
August 18th 2011, 09:09 PM
dagbayani481

Re: Trigonometric Equation

Quote:

Originally Posted by Prove It

Surely $\displaystyle x = 2\pi$ is a solution as well :)

Omg, facepalm. I could've sworn 2pi wasn't included in the interval. That's kind of dumb for them to include it considering 0 is basically the same as 2pi! Nonetheless, thank you for bringing that to my attention, I'll pay better attention to the problem next time :)
August 19th 2011, 03:17 AM
skeeter

Re: Trigonometric Equation

Quote:

Originally Posted by dagbayani481

So the questions asks find all the values of x in interval [0,2pi) ...

according to the instructions, $2\pi$ is not included in the interval

?
August 19th 2011, 03:21 AM
mr fantastic

Re: Trigonometric Equation

Quote:

Originally Posted by skeeter

according to the instructions, $2\pi$ is not included in the interval

?

Indeed. Yet on the screen shot in post #3 it's $[0, 2 \pi]$ ....?