Express $\cosh 2x$ and $\sinh 2x$ in exponential form and hence solve for real values of $x$ the equation:$2 \cosh 2x - \sinh 2x =2$ 
Express $\cosh 2x$ and $\sinh 2x$ in exponential form and hence solve for real values of $x$ the equation:
  $2 \cosh 2x - \sinh 2x =2$

Here is my idea:
$$2 \cosh 2x- \sinh 2x  = \frac{2(e^x)^2+2(e^{-x})^2}{2} - \frac{(e^x)^2-(e^{-x})^2}{2}$$
$$2=\frac{(e^x)^2+3(e^{-x})^2}{2}$$
$$4 = (e^x)^2 + 3(e^{-x})^2$$
$$4= (e^x)^2 + \frac{(3)(1)}{(e^x)^2 }$$
Multiplying both sides by $(e^x)^2$
$$4(e^x)^2= (e^x)^4 +3$$
or
$$(e^x)^4 - 4(e^x)^2 = -3$$
This sort of looks like something I could solve by completing the square or some other technique for solving a quadratic. But this is where I am stuck.
I know that $x=0$ and $x=0.549$ are the solutions.

Hint: i have $$e^{2x}+e^{-2x}-1/2e^{2x}+1/2e^{-2x}=2$$ multiplying by $e^{2x}$ we obtain $$e^{4x}+1-1/2e^{4x}+1/2=2e^{2x}$$ thus we get $$e^{4x}-4e^{2x}+3=0$$
with $$u=e^{2x}$$ we get $$u^2-4u+3=0$$
$u_1=3$ or $u_2=1$
