Answer
$y=-\frac{3}{\pi}x+2$
The graph of the curve and the tangent is below.
Work Step by Step
Find the parameter $t$ corresponding to the point $(0,2)$:
$(x,y)=(0,2)$
$(\sin\pi t,t^2+t)=(0,2)$
$\sin\pi t=0$ and $t^2+t=2$
$\sin \pi t=0$ and $t^2+t-2=0$
$\sin\pi t=0$ and $(t+2)(t-1)=0$
$t=0,\pm 1,\pm 2,\ldots$ and $t=-2,1$
$t=-2,1$
Find the slope of the tangent to the curve at $t=-2$ and $t=1$:
$m=\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t+1}{\pi \cos\pi t}$
When $t=-2$, $m=\frac{2\cdot (-2)+1}{\pi \cos(-2\pi)}=\frac{-3}{pi}=-\frac{3}{\pi}$
When $t=1$, $m=\frac{2\cdot 1+1}{\pi \cos(\pi)}=\frac{3}{-pi}=-\frac{3}{\pi}$
Find the equation of the tangent to the curve at $(0,2)$:
$y-2=m(x-0)$
$y-2=-\frac{3}{\pi}x$
$y=-\frac{3}{\pi}x+2$
Here is the graph of the curve and the tangent.
