The equation of tangent to the curve $\mathrm{y}=\cos (x+\mathrm{y})$ where $-2 \pi \leq x \leq 2 \pi$ and which is parallel to the line $x+2 \mathrm{y}=0$, is

The tangent to $y = \cos(x + y)$ parallel to $x + 2y = 0$ must share the line's slope, which by rearrangement to $y = -\frac{1}{2}x$ is $m = -\frac{1}{2}$.

Implicit differentiation of the curve yields:

$\frac{dy}{dx} = -\sin(x + y) \cdot (1 + \frac{dy}{dx})$

Solving for $\frac{dy}{dx}$ gives:

$\frac{dy}{dx} = -\frac{\sin(x + y)}{1 + \sin(x + y)}$

Equating this to $-\frac{1}{2}$ and simplifying leads to $\sin(x + y) = 1$, which implies $x + y = \frac{\pi}{2} + 2n\pi$ for integer $n$.

Substituting into the original equation and using $\cos(x + y) = 0$ gives $y = 0$, so $x = \frac{\pi}{2} + 2n\pi$.

Within $-2\pi \leq x \leq 2\pi$, valid points are $\left(\frac{\pi}{2}, 0\right)$ and $\left(-\frac{3\pi}{2}, 0\right)$.

The tangent line equation with slope $-\frac{1}{2}$ through $\left(\frac{\pi}{2}, 0\right)$ is:

$2x + 4y - \pi = 0$

This corresponds to option B. The other tangent does not match any given option.