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

The derivative is \(\frac{dy}{dx} = 1 - 2\\sin 2x\).

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

\(1 - 2\\sin 2x = 0\)

\(2\\sin 2x = 1\)

\(\\sin 2x = \frac{1}{2}\)

Solving for \(x\), we have:

\(2x = \frac{\pi}{6} + 2n\pi\) or \(2x = \frac{5\pi}{6} + 2n\pi\)

For \(0 \leq x \leq \pi\), the solutions are:

\(x = \frac{1}{12}\pi\) and \(x = \frac{5}{12}\pi\).

To determine the nature of these stationary points, we use the second derivative test.

The second derivative is \(\frac{d^2y}{dx^2} = -4\\cos 2x\).

For \(x = \frac{1}{12}\pi\):

\(\frac{d^2y}{dx^2} = -4\\cos \left(\frac{1}{6}\pi\right) = -4 \times \frac{\sqrt{3}}{2} = -2\sqrt{3} \lt 0\)

Thus, \(x = \frac{1}{12}\pi\) is a maximum point.

For \(x = \frac{5}{12}\pi\):

\(\frac{d^2y}{dx^2} = -4\\cos \left(\frac{5}{6}\pi\right) = -4 \times \left(-\frac{\sqrt{3}}{2}\right) = 2\sqrt{3} \gt 0\)

Thus, \(x = \frac{5}{12}\pi\) is a minimum point.