When does this parametric curve cross itself?

Question

Find the points where the curve given parametrically by$$\mathbf{r}(t)=\left(2+\cos\frac{3}{2}t\right)\left(\begin{matrix}\cos t\\\sin t\end{matrix}\right)$$crosses itself.

So, I understand that I should consider this equation$$\mathbf{r}(t_1)=\mathbf{r}(t_2),$$ which essentially is$$\left(2+\cos\frac{3}{2}t_1\right)\cos t_1=\left(2+\cos\frac{3}{2}t_2\right)\cos t_2$$and $$\left(2+\cos\frac{3}{2}t_1\right)\sin t_1=\left(2+\cos\frac{3}{2}t_2\right)\sin t_2,$$ but then I don't really know how to proceed.

Any help will be appreciated, especially if it's in the form of hints.

EDIT: I forgot to mention that $t\in [0,4\pi]$, which I guess simplifies things.

Answer 1

Hint: $2 + \cos x$ is never zero, so you can divide the first equation by the second. What do you get?

Answer 2

If you write down the parametric equation in polar coordinates instead of cartesian ones, you will immediately see that you need $t_1+2\pi = t_2$ (or vice versa) as well as $2+\cos \frac 32t_1 = 2+\cos\frac 32 t_2$. But $$\cos \frac 32t_2 = \cos(\frac 32 t_1 + 3\pi) = \cos(\frac 32 t_1 + \pi) = -\cos(\frac 32 t_1)$$ so $\cos\frac 32 t_1 = 0$ ...