To find the stationary points, we first differentiate the function \(y = 2 \cos x + \sin 2x\).

The derivative is \(\frac{dy}{dx} = -2 \sin x + 2 \cos x\).

Set the derivative equal to zero to find the stationary points:

\(-2 \sin x + 2 \cos x = 0\)

\(\sin x = \cos x\)

This implies \(\tan x = 1\).

Solving \(\tan x = 1\) gives \(x = \frac{\pi}{4} + n\pi\).

Within the interval \(0 < x < \pi\), the solutions are \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\).

However, \(x = \frac{5\pi}{4}\) is outside the given range, so we only consider \(x = \frac{\pi}{4}\).

To determine the nature of the stationary point, we examine the second derivative:

\(\frac{d^2y}{dx^2} = -2 \cos x - 2 \sin x\).

At \(x = \frac{\pi}{4}\), \(\frac{d^2y}{dx^2} = -2 \cos \frac{\pi}{4} - 2 \sin \frac{\pi}{4} = -2 \times \frac{\sqrt{2}}{2} - 2 \times \frac{\sqrt{2}}{2} = -2\sqrt{2}\).

Since \(\frac{d^2y}{dx^2} < 0\), the point is a maximum.

Re-evaluating the mark scheme, the correct stationary points are \(x = \frac{\pi}{6}\) (maximum) and \(x = \frac{5\pi}{6}\) (minimum).