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Math Help - Is there a systematic way to solve trig equations for all zero crossings?

  1. #1
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    Is there a systematic way to solve trig equations for all zero crossings?

    I'm taking far too long to solve these trig equations. I can't seem to find anything that gives me a systematic way to finding all solutions to a trig equation.

    For example, I have f(x) = \sin (2\pi x)

    It's not particularly difficult differentiate and solve this for x, and end up with:

    x = \frac {\cos^{-1}(0)}{2\pi} = \frac {1}{4}

    I can figure that another zero crossing occurs at -\frac {1}{4}, and observe that it is zero every \frac {1}{2}. But I find myself lacking a more systematic way of approaching these kinds of problems and solving them more efficiently that I do, currently.

    Another short example of one I did yesterday. f(x) = x + 2 \sin x \implies f'(x) = cos (2\pi x) \implies x = \cos^{-1}(-\frac {1}{2}). After much bashing of head against wall, I worked out that the solutions are all described by (1 + 2n)\pi \pm \frac {\pi}{3}~ for ~n \in \mathbb{Z}. But I can't help but think there's a better way to figure this stuff out. It took a lot of head scratching to figure that one out exactly.

    So is there some way I can more easily and quickly find every point at which trig equations make a zero crossing, no matter the period and such?

    Hope so, because they are becoming a bit of a pain.

    Any help or pointers to this information extremely appreciated, and would save me much time and annoyance.
    Last edited by Grep; April 21st 2009 at 09:03 AM.
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  2. #2
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    Hello Grep

    I think what you need is to study the following equations. They do provide general solutions to the sort of problems you've identified:

    For all integer values of n:

    • \sin A = \sin B \Rightarrow A = n\pi + (-1)^nB


    • \cos A = \cos B \Rightarrow A = 2n\pi \pm B


    Check them out for various values of n.

    The first one gives:

    n = 0: A = B

    n = 1: A = \pi - B

    n = 2: A = 2\pi + B

    n = -1: A = -\pi - B

    ... and so on.

    The second one is easier:

    n = 0: A = \pm B

    n = 1: A = 2\pi \pm B

    n = -1: A = -2\pi \pm B

    ... and so on.

    Does that help?

    Grandad
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  3. #3
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    It does help, sort of. The knowledge that there's a general solution makes me very happy.

    Just trying this out a bit, and not 100% sure I get it. But I'm actually really tired right now. Not the best time to figure things out... lol

    Let me see if I can apply this. Say, with this equation: f'(x) = \cos (2\pi x).

    Ok, so I have \cos A = \cos B \implies A = 2n \pi \pm B.

    So if I get this right, I solve for B and get \frac {1}{4}. So
    n = 0: A = 2(0)\pi \pm \frac {1}{4} = \pm \frac {1}{4}

    n = 1: A = 2(1)\pi \pm \frac {1}{4} = 2\pi \pm \frac {1}{4}
    etc...
    But I also know that \pm \frac {3}{4} is a solution, and I don't see that. So still a bit confused.

    Am I close, here? lol I really should get some sleep, then tackle this, but it's bugging me.

    Thanks a million for helping me, it's always very appreciated!

    Edit: Yeah, I really don't think I got it right. For one, B should be 2\pi x, I think.
    Last edited by Grep; April 21st 2009 at 08:55 AM.
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  4. #4
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    Eureka, I think I've got it.

    So cos (2\pi x) = cos (2\pi \frac {1}{4}) = cos (\frac {\pi}{2}). And \frac {\pi}{2} is my angle B.

    Thus,
    n = 0: x = \pm \frac {\frac {\pi}{2}}{2\pi} = \pm \frac {1}{4}

    n = 1: x = \frac {2\pi - \frac {\pi}{2}}{2\pi} = \frac {3}{4} and x = \frac {2\pi + \frac {\pi}{2}}{2\pi} = \frac {5}{4}

    Having to divide by 2\pi because my angle B was 2\pi x not just x, and I want to isolate the x.

    Thankyouthankyouthankyou!

    Now I can sleep... lol
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  5. #5
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    Excellent. You've got it!

    Grandad
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