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Math Help - Trigonometric Equation

  1. #1
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    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?
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  2. #2
    Super Member Quacky's Avatar
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    Re: Trigonometric Equation

    Quote Originally Posted by dagbayani481 View Post
    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?
    Last edited by Quacky; August 18th 2011 at 08:48 PM.
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  3. #3
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    Re: Trigonometric Equation

    Quote Originally Posted by Quacky View Post
    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:

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  4. #4
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    Re: Trigonometric Equation

    Surely \displaystyle x = 2\pi is a solution as well
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  5. #5
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    Re: Trigonometric Equation

    Quote Originally Posted by Prove It View Post
    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
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  6. #6
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    Re: Trigonometric Equation

    Quote Originally Posted by dagbayani481 View Post
    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

    ?
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  7. #7
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    Re: Trigonometric Equation

    Quote Originally Posted by skeeter View Post
    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] ....?
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