# Trigonometric Equation

• Aug 18th 2011, 05:54 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?
• Aug 18th 2011, 08:34 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?
• Aug 18th 2011, 09:40 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:

• Aug 18th 2011, 09:52 PM
Prove It
Re: Trigonometric Equation
Surely $\displaystyle x = 2\pi$ is a solution as well :)
• Aug 18th 2011, 10:09 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 :)
• Aug 19th 2011, 04:17 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

?
• Aug 19th 2011, 04:21 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 $[0, 2 \pi]$ ....?