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Mathematics 20 Online
OpenStudy (anonymous):

find the solution of 3sinhx-coshx=1

OpenStudy (kirbykirby):

Convert \(\sinh x\) and \(\cosh x\) into their exponential definitions:\[\large \begin{align} 3\sinh x- \cosh x&= 3\left(\frac{e^x-e^{-x}}{2}\right)-\left(\frac{e^x+e^{-x}}{2}\right)=1 \\ ~ \\ &=\frac{3}{2}e^x-\frac{3}{2}e^{-x}-\frac{1}{2}e^x-\frac{1}{2}e^{-x}=1\\ ~ \\&=e^x-2e^{-x}=1\\ \, \\ &\text{Let } \large e^x=y \\ ~ \\ &=y-2y^{-1}=1 \\ ~ \\ & \implies y - 2\frac{1}{y}-1=0 \\ \, \\ &\text{multiply both sides by }y \\ ~ \\ &\implies y^2-2-y=0\end{align} \] You now have a quadratic in \(y\). You can factor it and solve for \(y\). Then you just have to remember that \(y = e^x\), so convert your answers into exponential form and solve for \(x\) ! You might have to check which solutions will be valid for an exponential function.

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