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I want to solve the following equation for real values of $x$, by substituting the exponential forms of the hyperbolic functions. \begin{equation} 2\cosh(2x) - \sinh(2x) = 2 \end{equation} If someone could help me I would be most grateful. Thanks in advance!

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    $\begingroup$ If you expand the hyperbolic functions into the exponential form, what equation do you get? $\endgroup$ – Daniel Fischer Jul 29 '15 at 10:36
  • $\begingroup$ \begin{equation} e^{2x} + 3e^{-2x} = 4 \end{equation} $\endgroup$ – Nikos Jul 29 '15 at 10:39
  • $\begingroup$ Using definition of hyperbolic functions, you will get the equation of $e^{2x}$. $\endgroup$ – GAVD Jul 29 '15 at 10:39
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    $\begingroup$ Now set $y = e^{2x}$ and solve for $y$. $\endgroup$ – Daniel Fischer Jul 29 '15 at 10:40
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Notice, the following formula $$\color{blue}{\cosh \theta=\frac{e^{\theta}+e^{-\theta}}{2}}$$ & $$\color{blue}{\sinh \theta=\frac{e^{\theta}-e^{-\theta}}{2}}$$ Now, we have $$2\cosh(2x)-\sinh(2x)=2$$ $$2\left(\frac{e^{2x}+e^{-2x}}{2}\right)-\left(\frac{e^{2x}-e^{-2x}}{2}\right)=2$$ $$\frac{2e^{2x}+2e^{-2x}-e^{2x}+e^{-2x}}{2}=2$$ $$\frac{e^{2x}+3e^{-2x}}{2}=2$$ $$e^{2x}+\frac{3}{e^{-2x}}=4$$ $$e^{4x}-4e^{2x}+3=0$$ we have $$(e^{2x}-1)(e^{2x}-3)=0$$ $$\text{if}\ e^{2x}-1=0\iff e^{2x}= 1 \iff 2x=\ln 1\iff \color{blue}{x=0} $$ $$\text{if}\ e^{2x}-3=0\iff e^{2x}=3 \iff 2x=\ln 3\iff \color{blue}{x=\frac{1}{2}\ln 3=\ln\sqrt{3}} $$

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$$2\cosh(2x)-\sinh(2x)=2\Longleftrightarrow$$ $$-2+2\cosh(2x)-\sinh(2x)=0\Longleftrightarrow$$ $$-2\sinh(x)(\cosh(x)-2\sinh(x))=0\Longleftrightarrow$$ $$\sinh(x)(\cosh(x)-2\sinh(x))=0\Longleftrightarrow$$ $$\sinh(x)=0\vee\cosh(x)-2\sinh(x)=0\Longleftrightarrow$$ $$\sinh(x)=0\vee\cosh(x)=2\sinh(x)\Longleftrightarrow$$ $$\sinh(x)=0\vee\coth(x)=2\Longleftrightarrow$$ $$\sinh(x)=0\vee x=\coth^{-1}(2)\Longleftrightarrow$$ $$x=\sinh^{-1}(0)\vee x=\coth^{-1}(2)\Longleftrightarrow$$ $$x=0\vee x=\coth^{-1}(2)\Longleftrightarrow$$ $$x=0\vee x=\frac{\ln(3)}{2}$$

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