Solve the equation ( hyperbolic functions)

**Steve Mckenna** · July 28th 2009, 10:35 PM

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?

**DeMath** · July 28th 2009, 11:28 PM

Originally Posted by Steve Mckenna

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.

$6\sinh \theta - 4\cosh \theta - 3 = 0 \Leftrightarrow 6 \cdot \frac{{{e^\theta } - {e^{ - \theta }}}}<br /> {2} - 4 \cdot \frac{{{e^\theta } + {e^{ - \theta }}}}<br /> {2} - 3 = 0 \Leftrightarrow$

$\Leftrightarrow 3\left( {{e^\theta } - {e^{ - \theta }}} \right) - 2\left( {{e^\theta } + {e^{ - \theta }}} \right) - 3 = 0 \Leftrightarrow {e^\theta } - \frac{5}{{{e^\theta }}} - 3 = 0 \Leftrightarrow$

$\Leftrightarrow {e^{2\theta }} - 3{e^\theta } - 5 = 0.$

${\text{Let }}{e^\theta } = t \Rightarrow {e^{2\theta }} = {t^2}.$

${t^2} - 3t - 5 = 0 \Rightarrow {t_{1,2}} = \frac{{3 \pm \sqrt {29} }}{2}.$

$\frac{{3 + \sqrt {29} }}{2} > 0,{\text{ }}\frac{{3 - \sqrt {29} }}{2} < 0.$

${e^\theta } = \frac{{3 + \sqrt {29} }}{2} \Leftrightarrow \theta = \ln \frac{{3 + \sqrt {29} }}{2}.$

**girdav** · July 28th 2009, 11:30 PM

I have the same equation.
Let $e^{\theta} = t$ and we have to solve $t-5\frac1t -3=0 \Leftrightarrow t^2-3t-5 = 0$ .

**songoku** · July 28th 2009, 11:30 PM

Hi steve

Yes, that's correct

EDIT :
@girdav : there's typo, 1 should be 5 ^^

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