Finding points of self intersection A curve is parameterised by $$\mathbf{r}(t)=((2+\cos 3t)\cos 2t, (2+\cos 3t)(\sin 2t)), t\in [0,2\pi ]. $$
How can I find the self intersection points? I set $\mathbf{r}(t_1)=\mathbf{r}(t_2)$ and then tried to solve the simultaneous equations but it doesn't seem to work out. Any pointers?
 A: Geometrically
Mathematica shows

Three point are
$$(1,1.7),(-2,0),(1,-1.7)$$
Analytically
(deleted)
A: Parametric representation 
$$\mathbf{r}(t)=((2+\cos 3t)\cos 2t, (2+\cos 3t)\sin 2t)), t\in [0,2\pi ]$$
can be replaced by this (complex) one
$$\mathbf{r}(t)=(2+\cos 3t)e^{2it} , t\in [0,2\pi ]$$
Thus, we are looking for values of $t_1$ and $t_2$, $t_1 \neq t_2$, such that:
$$\tag{1}(2+\cos 3t_1)e^{2it_1}=(2+\cos 3t_2)e^{2it_2}$$
Two complex numbers are equal if and only if their modules are equal and their arguments are equal (modulo $2 \pi$).
Remark:
Due to the fact that $2+\cos 3t>0$ for any value of $t$ shows that $2+\cos 3t_1$  and $2+\cos 3t_2$ are the modules of the LHS and RHS of (1), resp.
Thus, (1) is equivalent to:
$$\cases{\cos(3t_1)=\cos(3 t_2)\\2t_1=2t_2 \ modulo \ 2 \pi}$$
As $$cos(u)=cos(v) \ \iff \ u=\pm v+K 2\pi:$$
the previous conditions are equivalent to
$$\cases{3t_1= s 3 t_2+K 2\pi\\2t_1=2t_2+K' 2 \pi}$$
where $s=\pm1$ and $K,K'$ are integers.
$$\tag{2}\iff \ \ \cases{t_1= s t_2+K 2\pi/3\\t_1=t_2+K' \pi}$$
The cases where $s=1$ will not give double points (because they lead to $t_1=t_2+K''2\pi$). We can thus assume $s=-1$, i.e.,
$$\tag{3}\iff \ \ \cases{t_1= -t_2+K 2\pi/3 \ \ (a)\\t_1=t_2+K' \pi \ \ (b)}$$
Adding and substracting (3)(a) and (3)(b):
$$\tag{4}\cases{2t_1=K2\pi/3+K'\pi \ \ \iff \ \ t_1=K\pi/3+K'\pi/2\ \ (a)\\2t_2=K2\pi/3-K'\pi \ \ \iff \ \ t_2=K\pi/3-K'\pi/2 \ \ (b)}$$
with the same values of $K$ and $K'$ in (4)(a) and (4)(b).
For example, if $K=2$ and $K'=1$, $(t_1,t_2)=(7\pi/6,\pi/6)$, giving point $(1,\sqrt{3}).$
I leave you the task to consider the different other cases (by taking different possible cases for integers $K$ and $K'$, some of them being redundant). You will end up with the three following solutions (up to an exchange between $t_1$ and $t_2$, of course):
$$(t_1,t_2)=(7\pi/6,\pi/6), \ \ (5\pi/6,11\pi/6), \ \   \ \ (\pi/2,3\pi/2), $$ yielding double points :
$$(1,\sqrt{3}),  \ \ (1,-\sqrt{3}), \ \ (-2,0)$$
respectively.
