6 sinh (theta) - 4 cosh (theta) - 3 = 0
giving the root correct to 3 decimal places.
I have arrived at the following equation thus far:
e^theta - 5e^-theta - 3 = 0.
Is this correct?
Yes, it's correct.
$\displaystyle 6\sinh \theta - 4\cosh \theta - 3 = 0 \Leftrightarrow 6 \cdot \frac{{{e^\theta } - {e^{ - \theta }}}}
{2} - 4 \cdot \frac{{{e^\theta } + {e^{ - \theta }}}}
{2} - 3 = 0 \Leftrightarrow$
$\displaystyle \Leftrightarrow 3\left( {{e^\theta } - {e^{ - \theta }}} \right) - 2\left( {{e^\theta } + {e^{ - \theta }}} \right) - 3 = 0 \Leftrightarrow {e^\theta } - \frac{5}{{{e^\theta }}} - 3 = 0 \Leftrightarrow$
$\displaystyle \Leftrightarrow {e^{2\theta }} - 3{e^\theta } - 5 = 0.$
$\displaystyle {\text{Let }}{e^\theta } = t \Rightarrow {e^{2\theta }} = {t^2}.$
$\displaystyle {t^2} - 3t - 5 = 0 \Rightarrow {t_{1,2}} = \frac{{3 \pm \sqrt {29} }}{2}.$
$\displaystyle \frac{{3 + \sqrt {29} }}{2} > 0,{\text{ }}\frac{{3 - \sqrt {29} }}{2} < 0.$
$\displaystyle {e^\theta } = \frac{{3 + \sqrt {29} }}{2} \Leftrightarrow \theta = \ln \frac{{3 + \sqrt {29} }}{2}.$