For the orbit of point $P$ described by the parametric equations $x(t)=\sin(nt)$ and $y(t)=\sin(mt)$, determine $n$ and $m$ so that point $P$ is in the origin exactly twice. EDIT: I believe range of values was $-2\pi<t<2\pi$
This was a question on a test I recently had, but I had no idea how to even start to answer it. Any help would be appreciated.
Unfortunately, without the exact formulation my answer is just a hint. You know that at some time $\tau$ you have $x(\tau)=y(\tau)=0$. The solutions for these equations are $$nt=i\pi\\mt=j\pi$$with $i,j\in\mathbb Z$. One solution is obviously $t=0$, corresponding to $i=j=0$. You also know that $m=n=0$ is not a valid solution ($P$ is always at origin), but you can have any of those be $0$, if the other one is not. Without loss of generality, say $m\ne 0$. If $n=0$ all you need is that $mt$ is a multiple of $\pi$ exactly once in your time interval (minus the $0$ case). So if you are looking for a $t$ in the $(0,2\pi)$ interval, you get $m\in[1,2)$.
In case both $m$ and $n$ are non-zero, it gets a little more complicated. By dividing the above equations you get $$\frac nm=\frac ij$$but you still have the constraint of the time interval, and number of times $P$ is at origin. So you can have for example $x=0$ having a lot of solutions in the $(0,2\pi)$ interval, and similarly for $y=0$, but only one of them is common. Therefore you need to know the exact formulation of the problem to be able to make these constraints.