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This seems to be a simple enough problem to find $x$, however there seems to be something missing $$f(x) = \sinh^2(x) - 2\cosh(x)$$ I know for a fact that there two $x$-intercepts for this function, as you can see here:

enter image description here

I tried using double angle formulas to change the terms into something easier to work with. This was just one of many approaches I tried, but failed at:

$\sinh^2(x) = \cosh(2x) - \cosh^2(x)$

$\cosh(2x) = 2 \cosh(x)^2 -1$

—> $2\cosh^2(x) - 1 - \cosh^2(x) - 2\cosh(x) = 0$.

And then I used a substitution for $\cosh(x)$ to find $x$, and I ended up with $$ x = \log\left(\sqrt{2}+1+\sqrt{2(\sqrt{2}+2)}\right) $$ as one of the answers, with the other $x$ value symmetric to it across the line $x = 0$. It was close, but incorrect. Would appreciate any help or guidance on what I should have been doing instead to get the answers I needed.

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    $\begingroup$ Done. See this link for math formatting $\endgroup$
    – EditPiAf
    Commented Apr 25, 2020 at 13:33
  • $\begingroup$ By the way, there's a quicker way to your equation with only $\cosh x$. Simply use $\cosh^2x-\sinh^2x=1$ to get $\sinh^2x=\cosh^2x-1$. $\endgroup$
    – Blue
    Commented Apr 25, 2020 at 14:22

2 Answers 2

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There's no need to go to $\cosh2x$: since $\sinh^2x=\cosh^2x-1$, the equation transforms into $$ \cosh^2x-2\cosh x-1=0 $$ so $\cosh x=1+\sqrt{2}$ (the negative root must be discarded). If $r=1+\sqrt{2}$, you have $$ e^{2x}-2re^x+1=0 $$ hence $$ e^x=r\pm\sqrt{r^2-1}=1+\sqrt{2}\pm\sqrt{2+2\sqrt{2}} $$ You know that the roots of the quadratic $t^2-2rt+1=0$ are reciprocal of one another, so the solutions are $$ x=\pm\log(1+\sqrt{2}+\sqrt{2+2\sqrt{2}}) $$

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  • $\begingroup$ Oh I see. Apparently I should have gotten the same answer with my method, but I made some arithmetic error along the way. But thanks for that, the substiution with r seemed to help make the calculations easier. $\endgroup$
    – TGamer
    Commented Apr 26, 2020 at 1:19
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From $\cosh^2(x) - 2\cosh(x)-1 = 0$, you get

$$\cosh x= 1+ \sqrt2$$

Then, use the identity $\cosh^{-1}t = \ln(t+\sqrt{t^2-1}) $ to obtain

$$x= \pm \cosh^{-1} (1+\sqrt2)=\pm\ln (1+\sqrt2+ \sqrt{2\sqrt2+2}) $$

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