Solve $3\cosh(x)+2\sinh(2x) =0$ When trying to solve $$3\cosh(x)+2\sinh(2x)=0$$
I have subbed in the definitions of the cosh and sinh functions:
$${\cosh x=\frac{e^{x}+e^{-x}}{2}}$$
$${\sinh x=\frac{e^{x}-e^{-x}}{2}}$$
Which has given me:
$$\frac{3e^{x}+3e^{-x}}{2} +e^{2x}-e^{-2x}=0$$
I can recognise that $(e^x)^2=e^{2x}$. I'm now thinking I should multiply through by $e^{2x}$ but I'm stuck at this point on how to proceed. I know $x=-\ln(2)$ is the solution, just not too sure how to get to it from here. 

Hint to continue $\textbf{your work}$: substitute $e^x=t$ to get $$3(t+\frac{1}{t})+2(t^2+\frac{1}{t^2})=0.$$ Then substitute $u=t+\frac{1}{t}$. Can you continue this?
