Answer
$y=\frac{1}{3}x+\frac{2\pi}{3}$
Work Step by Step
The equation of the tangent line to the graph of the curve $y(x)$ at the point $(a,y(a))$ is:
$$y=\frac{dy}{dx}|_{x=a,y=y(a)}(x-a)+y(a)$$
So the tangent line at the point $(\pi, \pi)$ is:
$$y=\frac{dy}{dx}|_{x=\pi,y=\pi}(x-\pi)+\pi$$
Using the implicit differentiating it follows:
$$(x+y)'\cos(x+y)=(2x-2y)'$$
$$(1+\frac{dy}{dx})\cos(x+y)=2-2\frac{dy}{dx}$$
$$\cos(x+y)+\frac{dy}{dx}\cos(x+y)=2-2\frac{dy}{dx}$$
$$\frac{dy}{dx}\cos(x+y)+2\frac{dy}{dx}=2-\cos(x+y)$$
$$\frac{dy}{dx}(\cos(x+y)+2)=2-\cos(x+y)$$
$$\frac{dy}{dx}=\frac{2-\cos(x+y)}{\cos(x+y)+2}$$
$$\frac{dy}{dx}_{x=\pi, y=\pi}=\frac{2-\cos(\pi+\pi)}{\cos(\pi+\pi)+2}=\frac{1}{3}$$
The equation of the tangent line is:
$$y=\frac{1}{3}(x-\pi)+\pi$$
$$y=\frac{1}{3}x-\frac{\pi}{3}+\pi$$
$$y=\frac{1}{3}x+\frac{2\pi}{3}$$
Therefore, the equation of the tangent line is:
$$y=\frac{1}{3}x+\frac{2\pi}{3}$$
