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Math Help - Another trigonometric equation problem

  1. #1
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    Another trigonometric equation problem

    The question:
    Find all solutions z \in C of cos(z) = cos(2)

    My attempt:
    cos(z) = cos(x)cosh(y) - isin(x)sinh(y) = cos(2)

    So we need:
    sin(x)sinh(y) = 0 [1]
    cos(x)cosh(y) = cos(2) [2]

    [1] is zero when x = k\pi or y = 0
    Substitute into [2]:
    cos(k\pi)cosh(0) \ne cos(2)

    Try just y = 0:
    So cos(x)cosh(0) = cos(x) which we want to equal cos(2), thus x = \pm2

    Try just x = k\pi
    So cos(k\pi)cosh(y) = \pm cosh(y) = cos(2)

    Case when k is odd:
    cosh(y) = -cos(2)

    Case when k is even:
    cosh(y) = cos(2)

    Therefore, y = cosh^-1(cos(2))

    So, z = \pm 2 + icosh^{-1}(cos(2))

    But the solution is \pm(2 + 2k\pi) :/

    What am I doing wrong? Thanks.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Glitch View Post
    Try just x = k\pi So cos(k\pi)cosh(y) = \pm cosh(y) = cos(2)

    We have

    x=k\pi \Rightarrow \cos (k\pi)\cosh y=(-1)^k \cosh y

    But

    \cosh y \geq 1\; (\forall y\in\mathbb{R})\textrm{\;and\;}\cos 2\neq \pm 1

    So, we only obtain solutions for y=0.
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