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

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

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