# Solve the following. (Hint: Ans is x=0 & x=0.5ln2) but how?

## $3 \cosh \left(2 x\right) = 3 + \sinh 2 x$

##### 1 Answer
Jan 26, 2018

The solutions are $S = \left\{0.5 \ln 2 , 0\right\}$

#### Explanation:

$\text{Reminder}$

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

$\sinh \left(x\right) = \frac{{e}^{x} - {e}^{-} x}{2}$

The equation is

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

$3 \cosh \left(2 x\right) - 3 - \sinh \left(2 x\right) = 0$

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

$3 \left(\left({e}^{2 x} + {e}^{- 2 x}\right)\right) - 6 - \left(\left({e}^{2 x} - {e}^{- 2 x}\right)\right) = 0$

$3 {e}^{2 x} + 3 {e}^{- 2 x} - 6 - {e}^{2 x} + {e}^{- 2 x} = 0$

$2 {e}^{2 x} + 4 {e}^{- 2 x} - 6 = 0$

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

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

${e}^{4 x} - 3 {e}^{2 x} + 2 = 0$

$\left({e}^{2 x} - 2\right) \left({e}^{2 x} - 1\right) = 0$

Therefore,

${e}^{2 x} - 2 = 0$, $\implies$, ${e}^{2 x} = 2$, $\implies$, $2 x = \ln 2$

$\implies$, $x = \frac{1}{2} \ln 2 = 0.5 \ln 2$

${e}^{2 x} - 1 = 0$, $\implies$, ${e}^{2 x} = 1$, $2 x = \ln 1 = 0$, $\implies$, $x = 0$