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.
$\endgroup$3 Answers
$\begingroup$HINT :
Let $(e^x)^2=e^{2x}=t$. Then, you'll have $$t^2-4t=-3$$
$\endgroup$ 1 $\begingroup$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 $u_2=1$
$\endgroup$ $\begingroup$Notice, $$(e^{x})^4-4(e^x)^2=-3$$ $$(e^{2x})^2-4(e^{2x})+3=0$$ $$(e^{2x})^2-3(e^{2x})-(e^{2x})+3=0$$ $$(e^{2x})((e^{2x})-3)-((e^{2x})-3)=0$$ $$(e^{2x}-1)((e^{2x})-3)=0$$ $$\implies e^{2x}-1=0\iff e^{2x}=1\iff 2x=\ln 1=0\iff x=0$$ & $$\implies e^{2x}-3=0\iff e^{2x}=3\iff 2x=\ln 3=0\iff x=\frac{1}{2}\ln 3=0.549$$
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