Looking for an equation that roughly generates the given graph

Question

I would like to precursor this with the fact that I am in no way a mathematician. Anyways, I am looking help finding an equation, series of equations, or any information on a good starting point for generating a graph roughly like the one in the picture below. Also, I'd like to be able to rotate it around 0,0. Any help is greatly appreciated.

Example of what I would roughly like to generate:

Answer 1

The parametric equations $$x=3sin(t-\frac{\pi}{4})+2sin(3t-\frac{\pi}{4})$$ $$y=3cos(t+\frac{\pi}{6})+2cos(3t+\frac{\pi}{6})$$ $$0≤t≤2\pi$$ gets you

I arrived at the numbers largely through trial and error, but started from this observation: from your sketch, if you look at the movement along just, say, the $x$-axis it describes a sine wave modulated by another sine. In other words, two sine functions added together. Using Geogebra, it was then a matter of playing with the various parameters to shift the sine curves with respect to one another until getting something approximating your sketch. No doubt there are more formal approaches.

Answer 2

The solution by @robert timmer-arends is good.

Here is a "a posteriori" possible full justification.

In fact one can relate the curve obtained to a general family, those of hypotrochoids, here given by:

$$z=5e^{\color{red}{1}it}-ae^{\color{red}{3}it} \ \ \iff \ \ \begin{cases}x&=&5\cos(t)-a\cos(3t)\\y&=&5\sin(t)-a\sin(3t) \end{cases}\tag{1}$$

represented on the left of the figure for values of $a$ ranging from $0$ to $5$. For the ratio $\color{red}{3/1}$, see remark below.

These curves can in a second step be "skewed" using a linear transform. Here (see the right part of the figure corresponding to $a=5$), we have taken:

$$\begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}\ \ 1&-\tfrac12\\-1& \ \ \ 1 \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\tag{2}$$

Plugging (1) into (2), no surprise that we find expressions close to those of Robert.

Remark: The ratio $\color{red}{3/1}$ in (1) is compulsory for having $3-1=2$ loops. Had we taken $z=5e^{\color{red}{1}it}-ae^{\color{red}{13}it}$, we would have generated $13-1=12$ loops.