Vì α là góc tù nên $\sin \alpha =\sqrt{1-{\cos }^2\alpha }$

Do đó: 

$\sin \alpha =-\cos \alpha =\frac{4}{5}$

$\iff \sqrt{1-{\cos }^2\alpha }-\cos \alpha =\frac{4}{5}$

$\iff \sqrt{1-{\cos }^2\alpha }=\frac{4}{5}+\cos \alpha \iff \left\{\begin{array}{c}1-{\cos }^2\alpha ={\left(\frac{4}{5}+\cos \alpha \right)}^2\\ \cos \alpha \ge -\frac{4}{5}\end{array}\right.$

$\begin{array}{c}50{\cos }^2\alpha +40\cos \alpha -9=0\\ \cos \alpha \ge -\frac{4}{5}\end{array}\iff \left\{\begin{array}{c}\left[\begin{array}{c}\cos \alpha =\frac{-4+\sqrt{34}}{10}\\ \cos \alpha =\frac{-4-\sqrt{34}}{10}\end{array}\iff \cos \alpha =\frac{-4-\sqrt{34}}{10}\right.\\ \cos \alpha \ge -\frac{4}{5}\end{array}\right.$(do α tù)

M = sinα – 2cosα = (sinα – cosα) – cosα =  $\frac{4}{5}-\frac{-4-\sqrt{34}}{10}=\frac{12+\sqrt{34}}{10}$.