Solve $3\cosh(x)+2\sinh(2x) =0$

Question

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.

Answer 1

Note that

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

Hence

$$3\cosh(x) + 4\sinh(x)\cosh(x) = 0$$

$$\cosh(x)\left(3 + 4\sinh(x)\right) = 0$$

Now it's easy.

Answer 2

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?

Answer 3

Starting from your last expression (I suppose a typo for $\sinh(2x)$ $$\frac{3e^{x}+3e^{-x}}{2} +e^{2x}-e^{-2x}=0$$ multiply everything by $2 e^{2x}$ to get $$-2+3 e^x+3 e^{3 x}+2 e^{4 x}=0$$ Let $t=e^x$ $$-2+3t+3t^3+2t^4=(t^2+1)(t+2)(2t-1)$$ There is only one possible solution in the real domain since $t=e^x$.