# Solving $\sinh^2x-2\cosh x = 0$

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:

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.

• Done. See this link for math formatting – EditPiAf Apr 25 '20 at 13:33
• By the way, there's a quicker way to your equation with only $\cosh x$. Simply use $\cosh^2x-\sinh^2x=1$ to get $\sinh^2x=\cosh^2x-1$. – Blue Apr 25 '20 at 14:22

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