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

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

First, find the derivatives: \(u' = -\sin x\) and \(v' = -2\sin 2x\).

Using the product rule: \(\frac{dy}{dx} = (-\sin x)(\cos 2x) + (\cos x)(-2\sin 2x)\).

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

Using the double angle formula \(\sin 2x = 2\sin x \cos x\), rewrite \(\cos 2x = 1 - 2\sin^2 x\).

Substitute and simplify: \(\frac{dy}{dx} = -\sin x (1 - 2\sin^2 x) - 4\cos^2 x \sin x\).

Factor out \(\sin x\): \(\frac{dy}{dx} = -\sin x (1 - 2\sin^2 x + 4\cos^2 x)\).

Set \(\frac{dy}{dx} = 0\): \(-\sin x (1 - 2\sin^2 x + 4\cos^2 x) = 0\).

Since \(\sin x = 0\) is not in the interval, solve \(1 - 2\sin^2 x + 4\cos^2 x = 0\).

Using \(\cos^2 x = 1 - \sin^2 x\), substitute: \(1 - 2\sin^2 x + 4(1 - \sin^2 x) = 0\).

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

Combine terms: \(5 - 6\sin^2 x = 0\).

Solve for \(\sin^2 x\): \(6\sin^2 x = 5\), so \(\sin^2 x = \frac{5}{6}\).

\(\sin x = \sqrt{\frac{5}{6}}\) or \(\sin x = -\sqrt{\frac{5}{6}}\).

Since \(0 < x < \frac{1}{2}\pi\), \(\sin x = \sqrt{\frac{5}{6}}\).

Find \(x\): \(x = \arcsin(\sqrt{\frac{5}{6}}) \approx 1.15\).