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.
Best Answer
HINT :
Let $(e^x)^2=e^{2x}=t$. Then, you'll have $$t^2-4t=-3$$