Possible $x$ values for a parametric equation $x=\sin t $ and $y=3\cos 2t$ over the interval $-\pi/2 \leq t \leq \pi/2$.
I know how to eliminate $t$, but I was asked to determine the possible $x$ values for the parametric equation and the Cartesian equation. I do not understand what the question is asking (indicate the portion of the graph traced by the particle). Am I supposed to plug all the values in the interval and see what values $x$ are the same?
 A: I suspect that the point is this: 
If you eliminate $t$, you'll get some implicit equation $f(x,y) = 0$. This equation defines a set $A = \{(x,y) \in \mathbb R^2 : f(x,y) = 0\}$.
The parametric equations generate another set $B = \{ (\sin t, 3\cos 2t) : -\pi/2 \le t \le \pi/2\}$.
The point of the exercise is to see that the sets $A$ and $B$ are not the the same. In fact $B$ is a proper subset of $A$. Saying it another way, the given paramerization does not "cover" the entire curve. Looking at the possible range of $x$ values generated by the parametric equations is one way to see this.
Specifically, $\sin t \le 1$ for $-\pi/2 \le t \le \pi/2$, so the parametric equations will only generate points $(x,y)$ with $x \le 1$ -- the moving particle never travels to the right of the vertical line $x = 1$. The point set $A$ described by the implicit equation does not have this restriction.
A: Since it's unclear what you want, I'd just guess what you are looking for.
If $-\pi/2 \leq t \leq \pi/2$ then $\sin^{-1}$ is defined and you can find $t$ by taking inverses:
$$ t= \sin^{-1}(x)$$
Now, plugging in $t$ you'll get:
$y=3\cos(2\sin^{-1}(x))=3((\cos(\sin^{-1}(x)))^2-(\sin(\sin^{-1}(x)))^2)$
$$y=3((\cos(\sin^{-1}(x))^2-x^2)=3((\sqrt{1-x^2} )^2-x^2)=3(1-2x^2)$$
You can also get this by using the trig identity $\cos(2t)=1-2\sin^2(t)$.
Now, notice that $ |x=\sin(t)| \leq 1$, therefore 
$|x|^2 \leq 1 \implies |x^2| \leq 1 \implies 0 \leq x^2 \leq 1 $
$ \implies -2 \leq -2x^2 \leq 0 \implies -1 \leq 1-2x^2 \leq 1$
$ \implies -3 \leq 3(1-2x^2) \leq 3 \implies -3 \leq y \leq 3 \implies |y| \leq 3$
Notice that $\sin(t)$ is continuous and it does get equal to $-1$ and $1$ at $-\pi/2$ and $\pi/2$ respectively. So, your parametrization only covers the part of the curve that lies inside the rectangle $(-1,1) \times (-3,3)$.
