7. Given $\alpha, \beta \geqslant 0, \alpha+\beta \leqslant 2 \pi$, then the minimum value of $\sin \alpha+2 \cos \beta$ is $\qquad$
If $\alpha \leqslant \pi$, then $\sin \alpha + 2 \cos \beta \geqslant 2 \cos \beta \geqslant -2$;
If $\alpha > \pi$, we only need to consider the case when $\beta > \frac{\pi}{2}$. Let $\alpha = \pi + \theta, \beta = \frac{\pi}{2} + \varphi, \theta + \varphi \leqslant \frac{\pi}{2}$, then $\sin \alpha + 2 \cos \beta = -\sin \theta - 2 \sin \varphi$. We now examine the maximum value of $\sin \theta + 2 \sin \varphi$. Since $\sin \theta, \sin \varphi$ are both monotonically increasing on $\left(0, \frac{\pi}{2}\right)$, the maximum value of $\sin \theta + 2 \sin \varphi$ is achieved when $\theta + \varphi = \frac{\pi}{2}$. At this point, $\sin \theta + 2 \sin \varphi = \cos \varphi + 2 \sin \varphi \leqslant \sqrt{5}$. Therefore, the minimum value of $\sin \alpha + 2 \cos \beta$ is $-\sqrt{5}$, with equality holding when $\sin \alpha = -\sin \theta = -\cos \varphi = -\frac{\sqrt{5}}{5}$, $\cos \beta = -\sin \varphi = -\frac{2 \sqrt{5}}{5}$.