# Given that cosh3x+3coshx=4cosh^3x how do you show that k^3 = 3k - 4 has a root in between -3 and -2 and choose a suitable value of y for k=ycoshx and calculate the root?

Dec 14, 2016

The real root is $k = - 2.195823345445647$

#### Explanation:

Making $k = y \cosh \left(x\right)$ and substituting into ${k}^{3} = 3 k - 4$ we have

${y}^{3} {\cosh}^{3} \left(x\right) = 3 y \cosh \left(x\right) - 4$

dividing both sides by $y \ne 0$ we have

${y}^{2} {\cosh}^{3} \left(x\right) x = 3 \cosh \left(x\right) - \frac{4}{y}$

making now $y = - 2$ and substituting we have

$4 {\cosh}^{3} \left(x\right) = 3 \cosh \left(x\right) + 2$

comparing now with the identity

$\cosh \left(3 x\right) + 3 \cosh \left(x\right) = 4 {\cosh}^{3} \left(x\right)$

we conclude

$\cosh \left(3 x\right) = 2$

Solving for $x$ we get $x = \pm 0.4389859656416055$

and finally

$k = - 2 \cosh \left(x\right) = - 2.195823345445647$ which is the sougth real root.

As we can observe

$- 3 < - 2.195823345445647 < - 2$