9.206. Arrange the three numbers $a_{1}=\log _{1} \sin 2 x$,

$$
a_{2}=-1-\log _{2} \sin x, a_{3}=\log _{\frac{1}{2}}(1-\cos 2 x), \text { if } 0<x<\frac{\pi}{4}
$$
## Solution.

Let's rewrite $a_{1}, a_{2}, a_{3}$ as follows:

$a_{1}=-\log _{2} 2 \sin x \cos x=-1-\log _{2} \sin x-\log _{2} \cos x$,

$a_{2}=-1-\log _{2} \sin x$,

$a_{3}=-\log _{2} 2 \sin ^{2} x=-1-2 \log _{2} \sin x$.

For $0<x<\frac{\pi}{4}, \log _{2} \sin x<0, \log _{2} \cos x<0$, therefore, the arrangement of these numbers in ascending order will be $a_{2}, a_{1}, a_{3}$.

Answer: $a_{2}, a_{1}, a_{3}$.

Solve the systems of inequalities $(9.207-9.214)$: