Answer
$$y=-x+\pi$$
Work Step by Step
First we will find the derivative:
$$y'=(\sin(\sin x))'=\cos(\sin x)(\sin x)'=\cos(\sin x)\cos x$$
The tangent line of the curve at the point $(x_{0},y_{0})$ is given by formula:
$$y-y_{0}=y'(x_{0})(x-x_{0}).$$
For $(x_{0},y_{0})=(\pi,0)$ we get:
$$y-0=\left.\cos(\sin x_{0})\cos x_{0}\right|_{x_{0}=\pi}(x-\pi)\Rightarrow y=\cos(\sin \pi)\cos\pi(x-\pi)\Rightarrow y=\cos0(-1)(x-\pi)\Rightarrow y=1\cdot(-1)(x-\pi)\Rightarrow y=-x+\pi$$
