Q: $\displaystyle 10 \mathrm{cosh} x + 2 \mathrm{sinh} x = 11$
Thanks in advance!
$\displaystyle 5 (e^x + e^{-x}) + (e^x - e^{-x}) = 11$
$\displaystyle \Rightarrow 6 e^x + 4 e^{-x} = 11$
$\displaystyle \Rightarrow 6 (e^x)^2 + 4 = 11 e^x$
$\displaystyle \Rightarrow 6 (e^x)^2 - 11 e^x + 4 = 0$
Substitute $\displaystyle u = e^x$:
$\displaystyle 6 u^2 - 11u + 4 = 0$.
Solve for u and equate the solutions to $\displaystyle e^x$. Solve for x.