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Solving hyperbolic trigonometric equations

  1. Oct 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that the real solution ##x## of $$tanhx=cosechx$$ can be written in the form ##x=ln(a \pm \sqrt{a})## and find an explicit value for ##a##.

    2. Relevant equations
    $$cosh^{2}x-sinh^{2}x=1$$
    $$coshx=\frac{e^{x}+e^{-x}}{2}$$

    3. The attempt at a solution
    I reduced the original equation $$tanhx=cosechx$$ to $$0=cosh^{2}x-coshx-1$$ I used the quadratic formula to get ##coshx= \frac{1 \pm \sqrt{5}}{2}##. I discarded ##\frac{1- \sqrt{5}}{2}## since ##coshx## must be greater than 0. I then expanded the solution to $$e^{x}+e^{-x}=1+ \sqrt{5}$$ I rearranged this to a quadratic in ##e^{x}## and solved using the quadratic formula and then took logs of both sides to get: $$x=ln\Bigg(\frac{1+ \sqrt{5}}{2} \pm \sqrt{\frac{1+ \sqrt{5}}{2}}\Bigg)$$ Obviously from this the solution can be expressed in the form ##x=ln(a \pm \sqrt{a})## when ##a= \frac{1+ \sqrt{5}}{2}##. I feel like this is cheating though since the question asks me to show the solution can be expressed in a specific form and then find the solution. Is it acceptable to find the solution and show it can be expressed in that form even with the wording of the question? I couldn't think of any other way to show the solution could be expressed in that form without explicitly solving it first.
     
  2. jcsd
  3. Oct 23, 2015 #2

    PeroK

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    Science Advisor
    Homework Helper
    Gold Member

    It's hardly cheating. I suspect it's just the wording of the question.
     
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