I have been trying to understand how to solve the exercise. I took a look at the answer and apparently you are supposed to solve $y(t) = 0$.
I do not understand this at all, I tried watching some videos on parametric equations but it just doesn‘t seem to correlate.
I mean, yes in a normal function I would understand if we were to equal the function to zero and plug the resulting $x$ into the original equation to find the y-coordinates. But why would we only search for $y(t)=0$ here and not $x(t)=0$
I would appreciate some tips on how to solve these, thanks in advance!
You have already been told $t=\frac{\pi}{2}$ is for $A$ so you just need to evaluate $x$ and $y$ there.
$$ x( \frac{\pi}{2} ) = \cos \pi - 3 \sin \frac{\pi}{2} - 1 = -1 - 3 -1 = -5\\ y( \frac{\pi}{2} ) = \sin \pi - 2 \cos \frac{\pi}{2} = 0 - 0 = 0 $$
So yes $A$ does intersect the $x$-axis because $y=0$ there. The x-axis has the equation $y=0$.
What is the other time for which $y(t)=0$
$$ y (t) = \sin 2 t - 2 \cos t $$
What about $\frac{3\pi}{2}$? It is similar to $\frac{\pi}{2}$ but on the other side.
$$ y (\frac{3\pi}{2}) = \sin (3 \pi) - 2 \cos \frac{3\pi}{2} = 0 - 0 $$
Good evaluate $x ( \frac{3\pi}{2}) = -1 - 3 (-1) - 1 = 1$