Answer
The slope is $\pi-1$.
Work Step by Step
Find $\frac{dy}{dt}$ and $\frac{dx}{dt}$:
$\frac{dx}{dt}=\frac{d}{dt}(t+\cos \pi t)=1-\pi\sin\pi t$
$\frac{dy}{dt}=\frac{d}{dt}(-t+\sin\pi t)=-1+\pi\cos \pi t$
Find $\frac{dy}{dx}$:
$\frac{dy}{dx}=\frac{-1+\pi\cos\pi t}{1-\pi \sin \pi t}$
Find the value of $t$ such that $(x,y)=(3,-2)$:
$t+\cos\pi t=3$ and $-t+\sin\pi t=-2$
$\cos\pi t=3-t$ and $\sin \pi t=t-2$ (Take the square to both)
$\cos^2\pi t=9-6t+t^2$ and $\sin^2\pi t=t^2-4t+4$ (Add both)
$\sin^2\pi t+\cos^2\pi t=2t^2-10t+13$ (Apply the Pythagorean Theorem)
$1=2t^2-10t+13$
$2t^2-10t+12=0$
$(2t-4)(t-3)=0$
$t=2\vee t=3$
For $t=2$,
$x=2+\cos2\pi=2+1=3$ and $y=-2+\sin 2\pi=-2$ (satisfied)
For $t=3$,
$x=3+\cos3\pi =3+(-1)=2$ and $y=-3+\sin3\pi =-3$ (not satisfied)
So, $t=2$.
Find the slope of the tangent to the curve at $(x,y)=(3,-2)$:
$\frac{dy}{dx}=\frac{-1+\pi\cos 2\pi}{1-\pi\sin 2\pi}=\frac{-1+\pi}{1-0}=\pi-1$
