If $$0 < x, y < \pi$$ and $$\cos x + \cos y - \cos(x + y) = \frac{3}{2}$$, then $$\sin x + \cos y$$ is equal to:

We are given $$\cos x + \cos y - \cos(x + y) = \frac{3}{2}$$ with $$0 < x, y < \pi$$.

Using the identities $$\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}$$ and $$\cos(x+y) = 2\cos^2\frac{x+y}{2} - 1$$, we substitute to get $$2\cos\frac{x+y}{2}\cos\frac{x-y}{2} - 2\cos^2\frac{x+y}{2} + 1 = \frac{3}{2}$$.

Let $$u = \cos\frac{x+y}{2}$$ and $$v = \cos\frac{x-y}{2}$$. Then $$2uv - 2u^2 = \frac{1}{2}$$, or equivalently $$4u^2 - 4uv + 1 = 0$$.

Treating this as a quadratic in $$u$$: $$4u^2 - 4vu + 1 = 0$$. The discriminant must be non-negative: $$16v^2 - 16 \geq 0$$, so $$v^2 \geq 1$$. Since $$v = \cos\frac{x-y}{2}$$ and $$|v| \leq 1$$, we need $$v = \pm 1$$. Given the constraint $$0 < x, y < \pi$$, we have $$\left|\frac{x-y}{2}\right| < \frac{\pi}{2}$$, so $$v > 0$$, giving $$v = 1$$, which means $$x = y$$.

With $$x = y$$, the equation becomes $$2\cos x - \cos 2x = \frac{3}{2}$$, i.e., $$2\cos x - (2\cos^2 x - 1) = \frac{3}{2}$$, which simplifies to $$2\cos^2 x - 2\cos x + \frac{1}{2} = 0$$, or $$4\cos^2 x - 4\cos x + 1 = 0$$.

This gives $$(2\cos x - 1)^2 = 0$$, so $$\cos x = \frac{1}{2}$$, meaning $$x = \frac{\pi}{3}$$. Therefore $$x = y = \frac{\pi}{3}$$.

Thus, $$\sin x + \cos y = \sin\frac{\pi}{3} + \cos\frac{\pi}{3} = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1 + \sqrt{3}}{2}$$.