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Solution:

Let $I=\int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}}\{2 \log \sin x-\log (2 \sin x \cos x)\} d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}}\{2 \log \sin x-\log \sin x-\log \cos x-\log 2\} d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}}\{\log \sin x-\log \cos x-\log 2\} d x$    ...(1)

It is known that, $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}}\{\log \cos x-\log \sin x-\log 2\} d x$    ...(2)

Adding (1) and (2), we obtain

$2 I=\int_{0}^{\frac{\pi}{2}}(-\log 2-\log 2) d x$

$\Rightarrow 2 I=-2 \log 2 \int_{0}^{\frac{\pi}{2}} 1 d x$

$\Rightarrow I=-\log 2\left[\frac{\pi}{2}\right]$

$\Rightarrow I=\frac{\pi}{2}(-\log 2)$

$\Rightarrow I=\frac{\pi}{2}\left[\log \frac{1}{2}\right]$

$\Rightarrow I=\frac{\pi}{2} \log \frac{1}{2}$