Determining the period of $ \frac{\sin(2x)}{\cos(3x)}$ I would like to compute the period of this function which is a fraction of two trigonometric functions. $$ \frac{\sin(2x)}{\cos(3x)}$$
Is there a theorem for this? what trick to use to easily find the period?
I started by reducing the fraction but I'm stuck on the rest.
For example, let $T$ be the period to be calculated: $$\frac{\sin(2x)}{\cos(3x)}  =\frac{\sin(2x + 2 T)}{\cos(3x+3T)} =  \frac{\sin(2x) \cos(2T)+\sin(2T) \cos(2x)}{\cos(3x) \cos(3T)-\sin(3T)\sin(3x)}$$
Thanks for your help.

Suppose the period is $p$, and suppose the domain of the function is suitably defined, then for all values of $x$ in the domain, we must have $$\frac{\sin 2x}{\cos 3x}=\frac{\sin(2(x+p))}{\cos(3(x+p))}$$
$$\implies \sin 2x\cos(3x+3p)=\cos 3x\sin(2x+2p)$$
If we set $x=0$, we have $$\sin(2p)=0\implies 2p=k\pi.k\in\mathbb{Z}$$
On the other hand, noting that $\sin 2x=\cos 3x$ when $x=\frac{\pi}{10}$, amongst other possible values, if we set $x=\frac{\pi}{10}$, we get $$\cos\left(\frac{3\pi}{10}+3p\right)=\sin\left(\frac{\pi}{5}+2p\right)=\cos\left(\frac{3\pi}{10}-2p\right)$$
$$\implies\frac{3\pi}{10}+3p=\pm\left(\frac{3\pi}{10}-2p\right)+n\cdot2\pi,n\in\mathbb{Z}$$
From this we get $p=n\cdot2\pi$ or $p=\frac{3\pi}{5}+n\cdot2\pi$
In order to satisfy this and the previous result for $p$ we have to choose $k$ and $n$ such that both equations $p=\frac{k\pi}{2}$ and $p=n\cdot2\pi$ are satisfied and $p$ has minimum value, so we choose $k=4$ and $n=1$.
Therefore the period is $2\pi$.
