Answer
See below:
Work Step by Step
Consider the provided function,
$ x=3\sin t $, $ y=4\cos t+2$
Consider the first function, $ x=3\sin t $
The interval is given from $ t=0$ to $ t=2\pi $.
Therefore, put values of $ t $ in the function as $0\le t\le 2\pi $
Consider second function, $ y=4\cos t+2$
The interval is given from $ t=0$ to $ t=2\pi $.
Therefore, put values of $ t $ in the function as $0\le t\le 2\pi $
For, $ t=\frac{\pi }{2}$
$\begin{align}
& x=3\sin t \\
& =3\cdot \sin \frac{\pi }{2} \\
& =3
\end{align}$
And,
$\begin{align}
& y=4\cos \left( t+2 \right) \\
& =4\cos \left( \frac{\pi }{2}+2 \right) \\
& =2
\end{align}$
For, $ t=\pi $
$\begin{align}
& x=3\sin t \\
& =3\cdot \sin \pi \\
& =0
\end{align}$
And,
$\begin{align}
& y=4\cos \left( t+2 \right) \\
& =4\cos \left( \pi +2 \right) \\
& =-2
\end{align}$
For, $ t=\frac{3\pi }{2}$
$\begin{align}
& x=3\sin t \\
& =3\cdot \sin \frac{3\pi }{2} \\
& =-3
\end{align}$
And,
$\begin{align}
& y=4\cos \left( t+2 \right) \\
& =4\cos \left( \frac{3\pi }{2}+2 \right) \\
& =2
\end{align}$
For, $ t=0\text{ or 2}\pi $
$\begin{align}
& x=3\sin t \\
& =3\cdot \sin 0 \\
& =0
\end{align}$
And,
$\begin{align}
& y=4\cos \left( t+2 \right) \\
& =4\cos \left( 0+2 \right) \\
& =6
\end{align}$
