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...
 A: $$2\cos(2x)+1=2(1-2\sin^2x)+1=3-4\sin^2x=\frac{3\sin(x)-4\sin^3x}{\sin x}=\frac{\sin(3x)}{\sin x}$$
Note: $\sin(x)\ne0$
A: Hint for LHS to RHS: Convert $\cos 2x$ to one of its three identities, then multiply and divide by $\sin x$.
Hint for RHS to LHS: Convert $\sin 3x$ to a known identity, then divide by $\sin x.$  Once you have that, add and subtract $1$ to get a common factor, then you will have a term in $\cos^2 x$ that can be converted into a double angle formula.
A: We have $$ 2\sin \alpha \cos \beta = \sin(\alpha+\beta)+\sin(\alpha-\beta)$$
so
$$ (2\cos 2x +1)\sin x = 2\sin x\cos 2x + \sin x= \sin 3x + \sin(-x) +\sin x = \sin 3x$$
so assuming that $\sin x\neq 0$
$$ 2\cos 2x +1 = {\sin 3x \over \sin x}$$
A: 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.
