\(f'(x) = 2\cos x - 2\sin 2x = 2\cos x(1 - 2\sin x)\)

\(f'(x) = 0\) on \((0, \pi)\): \(x = \frac{\pi}{2}\) (from \(\cos x = 0\)), and \(x = \frac{\pi}{6}, \frac{5\pi}{6}\) (from \(\sin x = \frac{1}{2}\)).

Values: \(f\left(\frac{\pi}{6}\right) = 1 + \frac{1}{2} = 1.5\), \(f\left(\frac{\pi}{2}\right) = 2 - 1 = 1\), \(f\left(\frac{5\pi}{6}\right) = 1.5\).

So there are local maxima at \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\), and a local minimum at \(\frac{\pi}{2}\).

Answer: (C) Both local minimum and local maximum