Step 1: Square-free route:
Write each side in the form $\sqrt2\sin(\cdot)$: $\cos x + \sin x = \sqrt2\sin(x + \frac{\pi}{4})$ and $\cos 2x + \sin 2x = \sqrt2\sin(2x + \frac{\pi}{4})$.
Step 2: Equate sines:
$\sin A = \sin B$ gives $A = B + 2n\pi$ or $A = \pi - B + 2n\pi$.
First: $x + \frac\pi4 = 2x + \frac\pi4 + 2n\pi$, so $x = -2n\pi$, that is $x = 2n\pi$.
Second: $x + \frac\pi4 = \pi - 2x - \frac\pi4 + 2n\pi$, so $3x = \frac{\pi}{2} + 2n\pi$, $x = \frac{2n\pi}{3} + \frac{\pi}{6}$.
Step 3: Read p and q:
$p = 2$, $q = 2$, so the ratio is 1:1.
Final Answer:
The ratio p:q is 1:1, option (A).
\[ \boxed{1:1} \]