Answer
$(y-\pi)=\frac{1}{3}(x-\pi)$
Work Step by Step
Let's start by using implicit differentiation.
$(sin(x+y)=2x-2y)'$
$(1+y')cos(x+y)=2-2y'$
Now, we'll plug in the given point, $(\pi, \pi)$
$(1+y')cos(\pi+\pi)=2-2y'$
Remember that $cos(2\pi)=1$, so
$1+y'=2-2y'$
$y'=\frac{1}{3}$, which is the slope of our tangent line. We plug this into point-slope form:
$(y-y_1)=m(x-x_1)$
$(y-\pi)=\frac{1}{3}(x-\pi)$
