6. 已知 $\cos \alpha+2 \sin \alpha=-\sqrt{5}$, 则 $\tan \alpha=$
由 $\left\{\begin{array}{l}\cos \alpha+2 \sin \alpha=-\sqrt{5}, \\ \sin ^{2} \alpha+\cos ^{2} \alpha=1,\end{array}\right.$ 得 $(\sqrt{5} \sin \alpha+2)^{2}=0, \therefore \sin \alpha=-\frac{2 \sqrt{5}}{5}, \cos \alpha=-\frac{\sqrt{5}}{5}, \therefore \tan \alpha=$