# Solving $4\sinh(2x)=\cosh(2x)$

Solve $$4\sinh(2x)=\cosh(2x)$$

So my method that I have used brings me to the answer of $$x=0$$ but this ins't correct and I cannot see what I've done wrong. My method is: $$4\sinh(2x)-\cosh(2x)=0$$ $$\frac{4(e^{2x}-e^{-2x})}{2}-\frac{e^{2x} + e^{-2x}}{2} =\frac{3e^{2x}-3e^{-2x}}{2}=0$$ $$3e^{2x}-3e^{-2x}=0$$ $$e^{2x}-e^{-2x}=0$$ $$e^{2x} - \frac{1}{e^{2x}}=0$$ $$e^{4x}-1=0$$ $$4x=\ln1$$ But $$\ln1=0$$ so I've come to a dead end. Any help would be great.

• There's an error when summing the two fractions.
– dfnu
Apr 10, 2019 at 12:06

Note that it should be $$\frac{4(e^{2x}-e^{-2x})}{2}-\frac{e^{2x} + e^{-2x}}{2} =\frac{3e^{2x}-5e^{-2x}}{2}=0\implies 3e^{4x}-5=0$$
Alternatively, as $$\cosh x>0$$, we can divide by $$\cosh 2x$$ so we solve $$\tanh2x=1/4$$ and the equation $$\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$ can be used.
Why don't you use the alternative fomulæ: $$\sinh u=\frac{\mathrm e^{2u}-1}{2\mathrm e^u}, \quad \cosh u=\frac{\mathrm e^{2u}+1}{2\mathrm e^u},$$ which yield instantly, setting $$t=\mathrm e^{4x}$$, the equation $$4(t-1)=t+1.$$