Evaluate the definite integral \( \int_{0}^{\pi/2} (2 \log \sin x - \log \sin 2x) \, dx \).

Rewrite \( \log \sin 2x = \log(2 \sin x \cos x) = \log 2 + \log \sin x + \log \cos x \). Then the integral becomes \( \int_{0}^{\pi/2} (\log \sin x - \log \cos x - \log 2) \, dx = 0 - \frac{\pi}{2}\log 2 = -\frac{\pi}{2}\log 2 \).