<s>[INST] 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.
 [/INST] $$\frac{\sin(2x)}{\cos(3x)}=\frac{2\sin(x)\cos(x)}{4\cos^3(x)-3\cos x}=\frac{2\sin(x)}{4\cos^2(x)-3}$$
$\sin x$ and $\cos x$ has a period of $2\pi$. Therefore $\frac{2\sin(x)}{4\cos^2(x)-3}$ has a period of $2\pi$.
The fundamental period must be $2\pi/n$ where $n\in\mathbb{N}$. We have
$$\frac{2\sin(x)}{4\cos^2(x)-3}=\frac{2\sin(x+2\pi/n)}{4\cos^2(x+2\pi/n)-3}$$
Substituting $x=\frac{(n-2)\pi}{n}$ we get$$\frac{2\sin(\frac{(n-2)\pi}{n})}{4\cos^2(\frac{(n-2)\pi}{n})-3}=\frac{2\sin(\pi)}{4\cos^2(\pi)-3}=0$$
$$\implies \sin\bigg(\frac{(n-2)\pi}{n}\bigg)=0$$
This is true only if $n=1$ or $2$. If $n=2$ then
$$\frac{2\sin(x)}{4\cos^2(x)-3}=\frac{2\sin(x+\pi)}{4\cos^2(x+\pi)-3}=-\frac{2\sin(x)}{4\cos^2(x)-3}$$
Therefore $n\ne2$. So the fundamental period is $2\pi$.
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