Finding where the parametric curve $(2\sin 2t, 3\sin t)$ crosses itself

Question

The parametric functions I am dealing with are: $x=2\sin2t$ and $y=3\sin t$

I know for a parametric graph to cross itself, there must be two distinct $t$, $t_1$ and $t_2$, that when placed into the two parametric functions, must produce the same ordered-pair; that is, $x=f(t_1)=f(t_2)$ and $y=g(t_1)=g(t_2)$.

With this, I have two system of equations:

(1) $2\sin 2t_1= 2\sin 2t_2 \implies 4\sin t_1\cos t_1=4\sin t_2\cos t_2 \implies \sin t_1=\dfrac{\sin t_2\cos t_2}{\cos t_1}$

(2) $3\sin t_1=3\sin t_2$

When I substituted $\sin t_1$ into (2), and simplified, I got $\cos t_2=\cos t_1$ Wouldn't that mean the graph intersects itself everywhere?

Answer 1

From equation (2), we can certainly conclude that we need $\sin t_1=\sin t_2$ if we're to have a hope of self-intersection, and from (1), we can conclude that we need $$\sin t_1\cos t_1=\sin t_2\cos t_2.$$ [Be careful about dividing by $\cos t_1$, as it might be $0$.] By substitution, then, we need $$\sin t_1\cos t_1=\sin t_1\cos t_2,$$ or equivalently, $$(\cos t_1-\cos t_2)\sin t_1=0.$$ Thus, there are two circumstances (not mutually exclusive) under which we have self-intersection:

(A) $\sin t_1=\sin t_2$ and $\cos t_1=\cos t_2$ (which happens if and only if $t_1$ and $t_2$ differ by an integer multiple of $2\pi$).

(B) $\sin t_1=\sin t_2=0$ (which happens if and only if $t_1$ and $t_2$ are both integer multiples of $\pi$).