若 $\sin \alpha+\sin \beta=\frac{\sqrt{3}}{3}(\cos \beta-\cos \alpha)$ 且 $\alpha \in(0, \pi), \beta \in(0, \pi)$, 则 $\alpha-\beta$ 等于 ()
$\because \alpha, \beta \in(0, \pi), \therefore \sin \alpha+\sin \beta>0 . \therefore \cos \beta-\cos \alpha>0, \cos \beta>\cos \alpha$, 又在 $(0, \pi)$ 上, $y=\cos x$是减函数. $\therefore \beta<\alpha \therefore 0<\alpha-\beta<\pi$

由原式可知: $2 \sin \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}=\frac{\sqrt{3}}{3}\left(-2 \sin \frac{\alpha+\beta}{2} \cdot \sin \frac{\beta-\alpha}{2}\right), \therefore \tan \frac{\alpha-\beta}{2}=\sqrt{3}, \therefore \frac{\alpha-\beta}{2}$ $=\frac{\pi}{3}, \therefore \alpha-\beta=\frac{2 \pi}{3}$.