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Math Help - Solving Trig Equation

  1. #1
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    Solving Trig Equation

    Hi all,

    To start with - the problem.

    3\sin^2\theta + 7\cos\theta = 5 solve for 0 <= \theta <= 2\pi

    The working I've done so far.

    1. 3(1 - \cos^2\theta) + 7\cos\theta = 5

    2. 3 - 3\cos^2\theta + 7\cos\theta = 5

    3. \cos\theta (7 - 3\cos\theta) = 2

    4. \cos\theta = 2 or (7 - 3\cos\theta) = 2

    5. 7 - 3\cos\theta = 2

    6. -3\cos\theta = -5

    7. \cos\theta = \frac{3}{5}

    Therefore, \theta = 53.13 deg or \theta = 306.87 deg

    Which means that \theta = \frac{53.13\pi}{180} or \theta = \frac{306.87\pi}{180}

    I'm just not sure if step 4 is correct. Can I simply ignore one of the options because \cos\theta = 2 is invalid?

    Cheers.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Step 4 is correct.
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  3. #3
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    Thanks How about the rest of it lol.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    I'm sorry; step 4 is incorrect!

    How does step 4 follow from step 3.

    3. \cos\theta (7 - 3\cos\theta) = 2

    How did you get to step 4?

    4. \cos\theta = 2 or (7 - 3\cos\theta) = 2

    This conclusion can be made only if the RHS=0. I think you're missing something.
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  5. #5
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    I was thinking along the lines of factorising.

    I.e.

    (x^2 + 3x) = 0
    x (x + 3) = 0
    x = 0 or (x + 3) = 0
    x = 0 or x = -3

    Edit - Posted before your addition, I can only do that when x = 0 ? Learnt something new (although I should have known that before I guess)

    Any further idea's of where to look?
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  6. #6
    MHF Contributor alexmahone's Avatar
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    3cos^2 \theta - 7cos \theta+2=0
    (3cos^2 \theta-6cos \theta - cos \theta+2)=0
    3cos \theta (cos\theta-2) - (cos \theta-2)=0
    (cos \theta-2) (3cos \theta-1)=0
    (cos \theta-2)=0 or (3cos \theta-1)=0
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  7. #7
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    Quote Originally Posted by Peleus View Post
    Hi all,

    To start with - the problem.

    3\sin^2\theta + 7\cos\theta = 5 solve for 0 <= \theta <= 2\pi

    The working I've done so far.

    1. 3(1 - \cos^2\theta) + 7\cos\theta = 5

    2. 3 - 3\cos^2\theta + 7\cos\theta = 5

    3. \cos\theta (7 - 3\cos\theta) = 2

    4. \cos\theta = 2 or (7 - 3\cos\theta) = 2

    5. 7 - 3\cos\theta = 2

    6. -3\cos\theta = -5

    7. \cos\theta = \frac{3}{5}

    Therefore, \theta = 53.13 deg or \theta = 306.87 deg

    Which means that \theta = \frac{53.13\pi}{180} or \theta = \frac{306.87\pi}{180}

    I'm just not sure if step 4 is correct. Can I simply ignore one of the options because \cos\theta = 2 is invalid?

    Cheers.
    you can't do that on step three you need to do the following:

    3-3cos^2x+7cosx=5

    take everything to the right hand side
    3cos^2x-7cosx+5-3=0
    3cos^2x-7cosx+2=0

    now you have to factorise
    i would replace the cos with y first
    3y^2-7y+2=0
    (3y-1)(y-2)=0

    so
    3cosx=1
    cosx=1/3

    or cosx=2
    x=error

    so x=70.53 or 289.47 (2dp)

    and i have used x instead of theta
    Last edited by scoobydoo4; May 27th 2009 at 02:35 PM.
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