Solve inequality $\sinh (2x) - 3 \sinh x \ > 0$ 
Solve inequality $\sinh (2x) - 3 \sinh x \ > 0$

Using hyperbolic trigo indenties:
$ 2 \sinh x \cosh x - 3 \sinh x \ > 0$
$\sinh x (2 \cosh x -3) \ > 0$
To solve this, we need to take

*

*$\sinh x (2 \cosh x -3) \ > 0$


*$\sinh x (2 \cosh x -3) \ < 0$
why is this the case ? I thought we are only solving for the inequality of $ > 0$
 A: Recall that in general

$$AB>0 \iff (A>0 \land B>0) \lor (A<0 \land B<0)$$

therefore to solve the given inequality
$$\sinh x (2 \cosh x -3) \ > 0$$
we need to solve separately two cases

*

*$\sinh x>0 \land 2 \cosh x -3  > 0$
or

*

*$\sinh x<0 \land 2 \cosh x -3  < 0$
and by the union of the solution for each case we can find the complete solution.
A: $\sinh x = \frac {e^x - e^{-x}}{2}$
$\sinh 2x - 3\sinh x = \frac {e^{2x} - 3e^{x} + 3e^{-x} - e^{-2x}}{2}$
Let $u = e^x$
$\frac {u^2 - 3u + u^{-1} + u{-2}}{2}$
Multiply numerator and denominator by $u^2$
$\frac {u^4 - 3u^3 + 3u - 1}{2u^2}$
Factor the numerator
$\frac {(u-1)(u+1)(u - \frac {3 - \sqrt {5}}{2})(u - \frac {3 + \sqrt {5}}{2})}{2u^2}$
$e^x > 0$ for all $x$ so $u> 0$ for all $x$
This gives us the intervals to consider $(0,\frac {3-\sqrt {5}}{2}),(\frac {3-\sqrt {5}}{2},1),(1,\frac {3+\sqrt {5}}{2},1),(\frac {3+\sqrt {5}}{2},\infty)$
The LHS is greater than $0$ when $u\in (\frac {3-\sqrt {5}}{2},1)\cup(\frac {3+\sqrt {5}}{2},\infty)$
or $x\in (\ln(\frac {3-\sqrt {5}}{2}),0)\cup(\ln(\frac {3+\sqrt {5}}{2}),\infty)$
Okay, for the work you have done....
$\sinh x(\cosh x - 3) > 0$
either $\sinh x > 0$ and $\cosh x > 3$
or $\sinh x < 0$ and $\cosh x < 3$
