To find the stationary point, we need to differentiate \(y = \\sin x \\cos 2x\) and set the derivative to zero.

Using the product rule, \(\frac{d}{dx}(uv) = u'v + uv'\), where \(u = \\sin x\) and \(v = \\cos 2x\).

Then, \(u' = \\cos x\) and \(v' = -2 \\sin 2x\).

The derivative is:

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

Using the double angle identity \(\\sin 2x = 2 \\sin x \\cos x\), the derivative becomes:

\(\frac{dy}{dx} = \\cos x \\cos 2x - 4 \\sin x \\cos x \\sin x\).

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

\(\\cos x \\cos 2x - 4 \\sin^2 x \\cos x = 0\).

Factor out \(\\cos x\):

\(\\cos x (\\cos 2x - 4 \\sin^2 x) = 0\).

Since \(\\cos x \neq 0\) in the interval \(0 < x < \frac{1}{2} \pi\), we have:

\(\\cos 2x - 4 \\sin^2 x = 0\).

Using the identity \(\\cos 2x = 1 - 2 \\sin^2 x\), substitute:

\(1 - 2 \\sin^2 x - 4 \\sin^2 x = 0\).

Simplify to get:

\(1 - 6 \\sin^2 x = 0\).

\(6 \\sin^2 x = 1\).

\(\\sin^2 x = \frac{1}{6}\).

\(\\sin x = \frac{1}{\sqrt{6}}\).

Solving for \(x\) in the interval \(0 < x < \frac{1}{2} \pi\), we find:

\(x \approx 0.421\).