To find the stationary points, we need to differentiate the function and set the derivative equal to zero.

The given function is \(y = 3 \cos 2x + 7 \sin x + 2\).

Differentiate with respect to \(x\):

\(\frac{dy}{dx} = -6 \sin 2x + 7 \cos x\).

Using the identity \(\sin 2x = 2 \sin x \cos x\), rewrite the derivative:

\(\frac{dy}{dx} = -6 (2 \sin x \cos x) + 7 \cos x\).

\(\frac{dy}{dx} = -12 \sin x \cos x + 7 \cos x\).

Factor out \(\cos x\):

\(\frac{dy}{dx} = \cos x (-12 \sin x + 7)\).

Set \(\frac{dy}{dx} = 0\):

\(\cos x (-12 \sin x + 7) = 0\).

This gives two equations:

1. \(\cos x = 0\)

2. \(-12 \sin x + 7 = 0\)

For \(\cos x = 0\), \(x = \frac{\pi}{2} = 1.57\).

For \(-12 \sin x + 7 = 0\):

\(\sin x = \frac{7}{12}\).

Solving for \(x\) in the interval \(0 \leq x \leq \pi\):

\(x = \arcsin \left( \frac{7}{12} \right) \approx 0.623\).

\(x = \pi - \arcsin \left( \frac{7}{12} \right) \approx 2.52\).

Thus, the \(x\)-coordinates of the stationary points are \(0.623, 1.57, 2.52\).