If an equation of a tangent to the curve, $$y = \cos(x + y)$$, $$- 1 \leq x \leq 1 + \pi$$, is $$x + 2y = k$$ then $$k$$ is equal to :

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

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

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

The slope of the given tangent is $$m = -1/2$$

$$-\frac{\sin(x+y)}{1 + \sin(x+y)} = -\frac{1}{2}$$

$$2\sin(x+y) = 1 + \sin(x+y) \implies \sin(x+y) = 1$$

When $$\sin(x+y) = 1$$, we know $$\cos(x+y) = 0$$.

Since the original curve is $$y = \cos(x+y)$$, we get $$y = 0$$.

Substituting $$y = 0$$ into $$\sin(x+y) = 1$$: $$\sin(x+0) = 1 \implies \sin x = 1 \implies x = \frac{\pi}{2}$$

$$\frac{\pi}{2} + 2(0) = k$$

$$k = \frac{\pi}{2}$$