How to go from $2\cos 2x+1$ to $\frac{\sin 3x}{ \sin x} $ In an exercice I stumbled upon this equation.

$\sin 3x(2\cos 2x+1)=\frac{\sin^23x}{\sin x}$

Or somewhat simplified :

$2\cos 2x+1 = \frac{\sin 3x}{ \sin x} $

I managed to  work out the right to the left , using a lot of formula's (starting with rewriting it as $\frac{\sin (2x + x)}{\sin x }$   and although it would be possible to reverse the steps to go from  left to right, I have the feeling that I am missing something trivial here (just started out to relearn trigonometry to help my kid out...), it's not obvious for me to see how to startout from left to right...

There's nothing wrong with what you did:
$$\frac{\sin(2x+x)}{\sin x} = \frac{\sin(2x)\cos x +\cos(2x)\sin x}{\sin x} $$
$$= \frac{2 \sin(x) \cos (x) \cos (x) + (2 \cos^2 x-1)\sin(x)}{\sin x}$$
$$= 2 \cos^2 x + 2 \cos^2 x - 1$$
$$= 2(2 \cos^2 x - 1) + 1 = 2 \cos 2x + 1$$
where $\cos(2x) = \cos^2 x - \sin^2 x = \cos^2 x - (1 - \cos^2 x) = 2 \cos^2 x - 1$, which can be memorised.
