1. Given $-2 \pi<\alpha<\beta<-\pi$. Then the range of $2 \alpha-\beta$ is $\qquad$
2. $(-3 \pi,-\pi)$.
From the given, we have $-4 \pi<2 \alpha<-2 \pi, \pi<-\beta<2 \pi$. Adding these, we get
$$
-3 \pi<2 \alpha-\beta<0 \text {. }
$$

But $\alpha<\beta$, so
$$
2 \alpha-\beta<2 \beta-\beta=\beta<-\pi \text {. }
$$

Combining (1) and (2), we get $-3 \pi<2 \alpha-\beta<-\pi$.
Analysis. A common mistake in solving this problem is to have only equation (1) without equation (2).