Coordinates where parametric equation intersects itself I want to show that the graph of the parametric equations $$x=3\cos^3(t), y =\sin(2t) $$ intersects itself at the origin $O$. 
Now I can do this by substitution, eg solving $0=3\cos^3t$ and $\sin2t = 0$ and showing two distinct values of $t$ satisfy this. I am trying to do it supposing I didn't know the answer was $O$. 
My attempt:
For the curve to intersect itself, require $a, b$  ($a \neq b$) such that $$3\cos^3 a = 3\cos^3b, \hspace{4mm} \sin2a = \sin2b$$
This gives $a= b + 2m\pi$ and $a= b + 2n\pi$. But I'm not sure how to proceed from here... 
 A: You say : $ a \ne b $ , but that should be : $ a \ne b \mod(2\pi)$ because when $a=b+2n\pi$ we do get the same point on the curve and we might say the curve intersects itself there in a trivial way.
From $\sin(2a)=\sin(2b)$ we have: $2a=2b  +2n\pi \lor 2a=\pi-2b +2n\pi $ $\implies a=b +(2n'+1)\pi  \lor a=-b +\frac{1}{2}\pi +n\pi $. 
The first possibility : $a=b +(2n'+1)\pi$ is written like this to ensure that the integer factor before $\pi$ is odd because the case : $a=b \mod(2\pi)$ is not allowed because of the remark above.
From $\cos^3(a)=\cos^3(b)$ we have: $ a=b +2m\pi \lor a=-b +2m\pi$ . But the first : $ a=b +2m\pi$ is not allowed because of the remark above. 
So we must have one of these two cases :
I) $a=-b +2m\pi \land a=b +(2n'+1)\pi \implies a = ((n'+m)+\frac{1}{2})\pi \implies 3\cos^3(a)=0 \,\,\land \,\, \sin(2a)=0$
or :
II) $a=-b +2m\pi \land a=-b +\frac{1}{2}\pi +n\pi  \implies 2m\pi = \frac{1}{2}\pi +n\pi $ , this is a contradiction.
So $(0,0)$ is the only point where the curve intersects itself.
A: Rearrange your equations,
$$3\cos^3 a = 3\cos^3b, \hspace{4mm} \sin2a = \sin2b$$
as
$$0=\cos^3 a - \cos^3b$$
$$=(\cos a - \cos b)(\cos^2 a + \cos a \cos b+ \cos^2 b)$$
$$=-2\sin\frac{a+b}{2}\sin\frac{a-b}{2}(\cos^2 a + \cos a \cos b+ \cos^2 b)$$
$$0=\sin2a - \sin2b=2\cos(a+b)\sin(a-b)$$
Examine the joint solutions
$$\sin\frac{a-b}{2}=0, \>\>\>\>\> \sin(a-b)=0$$
which are satisfied if
$$b=a+2k\pi$$
So, let $a=\frac\pi2$, which satisfies,
$$0=3\cos^3a,\>\>\>\>\>\sin2a = 0$$
Then, all $b$'s given by $b = \frac\pi2+2k\pi$ also satisfy,
$$0=3\cos^3b,\>\>\>\>\>\sin2b = 0$$
The above result shows that the curve initially intersects itself at origin at $\frac\pi2$ and repeats it every $2\pi$ afterwards.
