Theo bài ra:
\(\frac{\pi}{2}< a< \pi\) => a nằm ở góc phần tư thứ 2
=> sina > 0 ; cos a<0
Ta có:
\(\left\{{}\begin{matrix}sina+2cosa=-1\\sin^2a+cos^2a=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}sina=-1-2cosa\\\left(-1-2cosa\right)^2+cos^2a=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}sina=-1-2cosa\\1+4cosa+4cos^2a+cos^2a=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}sina=-1-2cosa\\5cos^2a+4cosa=0\end{matrix}\right.\)
Vì cosa < 0
\(\Rightarrow\left\{{}\begin{matrix}sina=-1-2cosa\\cosa=-\frac{4}{5}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}sina=\frac{3}{5}\\cosa=-\frac{4}{5}\end{matrix}\right.\)
\(sin2a=2sina.cosa=2.\frac{3}{5}.\frac{-4}{5}=-\frac{24}{25}\)
