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Fairly basic trigonometric equation

  1. Jan 4, 2008 #1
    Acos(A*x)-Bcos(B*x)=0, where A>B

    Is there a general solution for an equation of this form?

    Thanks,
    intangible
     
  2. jcsd
  3. Jan 4, 2008 #2

    Gib Z

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    I'm quite inclined to say there are no solutions for that :(
     
  4. Jan 5, 2008 #3
    Acos(A*x) = Bcos(B*x) -- So we conclude that A = B
     
  5. Jan 6, 2008 #4
    How about A=Pi/2

    B=3Pi/2
     
  6. Jan 6, 2008 #5

    Gib Z

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    I really can not be bothered to expand cos (3x), but if you are, I will verify the rest. I highly doubt it works though.
     
  7. Jan 6, 2008 #6
    Thanks to everyone participating, I think I got sorted it out by inspecting a similar behaving sine and its infinite product.

    [tex]\sin x = x \prod_{n = 1}^\infty\left(1 - \frac{x^2}{\pi^2 n^2}\right)[/tex]

    If we assume Asin(pi*A*x)=Bsin(pi*B*x) we may easily inspect the existence of trivial roots if there are any. It turns out there are: for every natural number A and B (A>B) there is a trivial root at x=n. Also, if there is a natural number D so that it divides both A and B, then there will be the trivial roots x=n/D.

    [tex]A \prod_{n=1}^\infty\left (A x - n)(A x + n)\right = B \prod_{n=1}^\infty\left (B x - n)(B x + n)\right[/tex]

    That leaves A-1 unknown roots per cycle if the former is the case, (A-B)/D per cycle for the latter. Unfortunately, the other roots clearly are irrational, perhaps even transcendent.

    Regards,
    intangible

    (How do I mark this thread solved?)
     
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