$\begin{array}{rl} & y=\cos (x+y) ;-1 \leq x \leq 1+\pi \\ \Rightarrow & \frac{d y}{d x}=-\{\sin (x+y)\} \cdot\left(1+\frac{d y}{d x}\right) \\ \Rightarrow & \frac{d y}{d x}(1+\sin (x+y))=-\sin (x+y) \\ \Rightarrow & \frac{d y}{d x}=\frac{-\sin (x+y)}{1+\sin (x+y)} \\ \Rightarrow & \frac{d y}{d x}=\operatorname{slope} \text { of tangent }=\frac{-1}{2}=\frac{-\sin (x+y)}{1+\sin (x+y)} \\ \therefore \quad & \sin (x+y)=1 \\ x+ & y=(2 n+1) \frac{\pi}{2} \\ \Rightarrow \quad & y=\cos (x+y)=\cos \left[(2 n+1) \frac{\pi}{2}+y-y\right] \\ \Rightarrow \quad & y=\cos (2 n+1) \frac{\pi}{2}=0 \\ \therefore \quad \text { At } n=0, x+y=\frac{\pi}{2} & x=\frac{\pi}{2} \\ \therefore \quad & \\ x+2 y=k & k=\frac{\pi}{2} .\end{array}$