I have to find the inflection points of the curve $$\alpha(t)=(\sin(2t),\cos(3t)),\quad-\pi\leq t\leq\pi$$
The exercise explains itself that a inflection point is such that $\alpha'(t)$ and $\alpha''(t)$ are parallels.
Since $$\alpha'(t)=(2\cos(2t),-3\sin(3t))$$ $$\alpha''(t)=(-4\sin(2t),-9\cos(3t))$$ then if $\alpha'$ and $\alpha''$ are parallels, the following cross product $$\begin{vmatrix} i & j & k \\ 2\cos(2t) & -3\sin(3t) & 1 \\ -4\sin(2t) & -9\cos(3t) & 1 \end{vmatrix}$$ will be $\vec{0}$. So $$\left\{\begin{array}{l} -3\sin(3t)+9\cos(3t)=0 \\ -4\sin(2t)-2\cos(2t)=0 \\ -18\cos(2t)\cos(3t)-12\sin(2t)\sin(3t)=0 \end{array}\right.$$
I solved this system using Wolphram which says there are no solutions. However, when I see $\alpha$ plot, for me, there are 4 inflection points.