Question

Solve the following equation giving your answers in terms of natural logarithms
$4\tan hx=1+\sec hx$

Answer 1

$4\tan hx-\sec hx=1$

$4\left(\frac{e^x-e-x}{e^x+e^{-x}}\right)+\frac{2}{e^x+e^{-x}}=1$

$3e^{2x}+2e^x-5=0$

$e^x=-\frac{5}{3}$ or $1$

$x=1$ only