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Thread: Solve the Following Trigonometric Equation

  1. #1
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    Solve the Following Trigonometric Equation

    Hey, in pre-cal we're going over some of the stuff we learned this semester. Unfortunately for me, I have forgotten a bit of it:

    So, here's the problem:
    Solve the equation:
    sin2x + cos3x = 0 for [0, 2pi)

    I wish I could just graph it, but I have to do it algebraically and show all work. Any clues as to how to approach this would be appreciated

    Thanks
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  2. #2
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    Quote Originally Posted by ayoYNr View Post
    Hey, in pre-cal we're going over some of the stuff we learned this semester. Unfortunately for me, I have forgotten a bit of it:

    So, here's the problem:
    Solve the equation:
    sin2x + cos3x = 0 for [0, 2pi)

    I wish I could just graph it, but I have to do it algebraically and show all work. Any clues as to how to approach this would be appreciated

    Thanks
    first task ... you'll need to expand $\displaystyle \sin(2x)$ and $\displaystyle \cos(3x)$

    $\displaystyle \sin(2x) = 2\sin{x}\cos{x}$ ... that's done.

    $\displaystyle \cos(3x) = \cos(2x+x) = \cos(2x)\cos{x} - \sin(2x)\sin{x} = $ ... keep going
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  3. #3
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    Ok, so now I have:

    2sinxcosx + cos(2x)cosx - sin(2x)sinx

    Where do I go from here?
    Last edited by ayoYNr; May 10th 2009 at 11:58 AM.
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    (Sorry, ignore this post)
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  5. #5
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    Quote Originally Posted by ayoYNr View Post
    Ok, so now I have:

    2sinxcosx + cos(2x)cosx - sin(2x)sinx

    Where do I go from here?
    $\displaystyle 2\sin{x}\cos{x} + (1 - 2\sin^2{x})\cos{x} - 2\sin^2{x}\cos{x} = 0$

    $\displaystyle \cos{x}(2\sin{x} + 1 - 2\sin^2{x} - 2\sin^2{x}) = 0$

    $\displaystyle \cos{x}(1 + 2\sin{x} - 4\sin^2{x}) = 0$

    $\displaystyle \cos{x} = 0$ ... should be straightforward to solve for x

    $\displaystyle 1 + 2\sin{x} - 4\sin^2{x} = 0$ ... you'll need to use the quadratic formula to solve for $\displaystyle \sin{x}$
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