Find the equation of all the tangents to the curve $y=\cos(x+y)$ for $-2\pi \leq x \leq 2\pi$ that are parallel to the line $x+2y=0$.

(A) Given the curve $y = \cos(x+y)$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\sin(x+y) \left(1 + \frac{dy}{dx}\right)$.
Rearranging,we have $\frac{dy}{dx} (1 + \sin(x+y)) = -\sin(x+y)$,so $\frac{dy}{dx} = -\frac{\sin(x+y)}{1 + \sin(x+y)}$.
Since the tangent is parallel to $x+2y=0$,its slope is $m = -\frac{1}{2}$.
Equating the derivative to the slope: $-\frac{\sin(x+y)}{1 + \sin(x+y)} = -\frac{1}{2} \Rightarrow 2\sin(x+y) = 1 + \sin(x+y) \Rightarrow \sin(x+y) = 1$.
Since $\sin(x+y) = 1$,we have $\cos(x+y) = 0$. Given $y = \cos(x+y)$,this implies $y = 0$.
Substituting $y=0$ into the original equation: $0 = \cos(x+0) \Rightarrow \cos(x) = 0$.
For $-2\pi \leq x \leq 2\pi$,the solutions are $x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}$.
Checking $\sin(x+y) = 1$: For $x = \frac{\pi}{2}, y=0$,$\sin(\frac{\pi}{2}) = 1$ (Valid). For $x = -\frac{3\pi}{2}, y=0$,$\sin(-\frac{3\pi}{2}) = 1$ (Valid). For $x = -\frac{\pi}{2}, y=0$,$\sin(-\frac{\pi}{2}) = -1$ (Invalid). For $x = \frac{3\pi}{2}, y=0$,$\sin(\frac{3\pi}{2}) = -1$ (Invalid).
The points of tangency are $(\frac{\pi}{2}, 0)$ and $(-\frac{3\pi}{2}, 0)$.
Equation of tangent at $(\frac{\pi}{2}, 0)$ with slope $-\frac{1}{2}$: $y - 0 = -\frac{1}{2}(x - \frac{\pi}{2}) \Rightarrow 2x + 4y - \pi = 0$.
Equation of tangent at $(-\frac{3\pi}{2}, 0)$ with slope $-\frac{1}{2}$: $y - 0 = -\frac{1}{2}(x + \frac{3\pi}{2}) \Rightarrow 2x + 4y + 3\pi = 0$.