Find the values of $\sinh 2x$ and $\cosh 2x$, if $\sinh 3x = 3/4$ First, use $\sinh 3x = 3/4$ to find the value of $\sinh x$.
Then, we know that
$$\cosh^2 x - \sinh^2 x = 1$$
By substituting the value for $\sinh x$ in this equation we can find the value for $\cosh x$.
Then use the $\cosh x$ and $\sinh x$ to find $\sinh 2x$ and $\cosh 2x$ by substituting in

*

*$ \sinh 2x = 2 \sinh x \cosh x$


*$ \cosh 2x = 2 \cosh^2x - 1 $
Right now I can't seem to find a way to obtain the $\sinh x$ value from the given $\sinh 3x$.
 A: The inverse function of $f(x)=\sinh x$ is defined as $\operatorname{arsinh}x$, and it is quite simple to show that $$\operatorname{arsinh} x=\ln(x+\sqrt{x^2+1})$$
To solve your equation then, since we have
$$\sinh 3x=\frac{3}{4}$$
that means that
$$3x=\operatorname{arsinh}\frac{3}{4}=\ln\left(\frac{3}{4}+\sqrt{\frac{9}{16}+1}\right)=\ln2$$
You should be able to do the rest yourself. If you do need more help please don't hesitate to ask :)

By definition, $$\sinh x=\frac{e^x-e^{-x}}{2}$$
so setting $x=\frac{1}{3}\ln2$, we can see that
$$\sinh x=\frac{e^{\frac{1}{3}\ln2}-e^{-\frac{1}{3}\ln2}}{2}=\frac{2^{1/3}-0.5^{1/3}}{2}$$

An attempt that fails:
You may be tempted to use the formula for $\sinh3x$, derived below, but that isn't very helpful I'm afraid.
We know that
$$\sinh(2x)\equiv 2\sinh x\cosh x\quad\cosh(2x)\equiv 1+2\sinh^2x$$
and more generally,
$$\sinh(A+B)\equiv\sinh A\cosh B+\sinh B\cosh A.$$
Here's the sneaky part. We can write $\sinh(3x)$ as $\sinh(2x+x)$ and then use the addition formula above!
$$\begin{align}\sinh(3x)&\equiv\sinh(2x+x)\equiv\sinh2x\cosh x+\sinh x\cosh2x\\
&\equiv2\sinh x\cosh^2x+\sinh x(1+2\sinh^2x)\\
&\equiv2\sinh x(1+\sinh^2x)+\sinh x +2\sinh^3x\\
&\equiv3\sinh x+4\sinh^3x\end{align}$$
So your equation is really
$$3\sinh x+4\sinh^3x=\frac{3}{4}$$
The problem is, solving this equation for $\sinh x$ is very hard without using the cubic formula.
