Step-by-Step Solution
Key Concept: Convert logarithmic equation to algebraic form, identify feasible domain constraints, and solve using trigonometric identities
From $\log_{\sqrt{2}+1}(1 + \cos x) = 2$, we get $1 + \cos x = (\sqrt{2}+1)^2$ and $\sqrt{2}\sin x = 1, \sqrt{2}\sin x > 0$ with $x \in (0, \pi)$. This gives feasible region $x \in (0, \frac{\pi}{2}]$. Squaring and solving: $2\sin^2 x = 1 - \cos x$ leads to $(2\cos x - 1)(\cos x + 1) = 0$, so $\cos x = \frac{1}{2}$ or $\cos x = -1$. Since $\cos x = \frac{1}{2}$ is valid ($x = \frac{\pi}{3}$) and $\cos x = -1$ is outside the feasible region, we get $p = 1, q = 3$, therefore $p^2 + q^2 = 10$.
Correct Answer: 10