Cho góc lượng giác α thỏa mãn s i n ( 2 α ) + s i n ( 5 α ) − s i n ( 3 α ) 2 c o s 2 ( 2 α ) + c o s ( α ) − 1 = − 2 . Tính sin ( α ) .

\[\frac{{{\rm{sin}}\left( {{\rm{2\alpha }}} \right){\rm{ + sin}}\left( {{\rm{5\alpha }}} \right) - {\rm{sin}}\left( {{\rm{3\alpha }}} \right)}}{{{\rm{2co}}{{\rm{s}}^{\rm{2}}}\left( {{\rm{2\alpha }}} \right){\rm{ + cos}}\left( {\rm{\alpha }} \right) - {\rm{1}}}}{\rm{ = }} - {\rm{2}}\]

\[ \Leftrightarrow \frac{{2\sin \left( \alpha \right)\cos \left( \alpha \right) + 2\cos \left( {4\alpha } \right)\sin \left( \alpha \right)}}{{2.\frac{{1 + \cos \left( {4\alpha } \right)}}{2} + \cos \left( \alpha \right) - 1}}{\rm{ = }} - 2\]

\[ \Leftrightarrow \frac{{{\rm{2sin}}\left( {\rm{\alpha }} \right){\rm{cos}}\left( {\rm{\alpha }} \right){\rm{ + 2cos}}\left( {{\rm{4\alpha }}} \right){\rm{sin}}\left( {\rm{\alpha }} \right)}}{{{\rm{cos}}\left( {{\rm{4\alpha }}} \right){\rm{ + cos}}\left( {\rm{\alpha }} \right)}}{\rm{ = }} - 2\]

\[ \Leftrightarrow \frac{{{\rm{2sin}}\left( {\rm{\alpha }} \right)\left[ {{\rm{cos}}\left( {\rm{\alpha }} \right){\rm{ + cos}}\left( {{\rm{4\alpha }}} \right)} \right]}}{{{\rm{cos}}\left( {{\rm{4\alpha }}} \right){\rm{ + cos}}\left( {\rm{\alpha }} \right)}}{\rm{ = }} - 2\]

\[ \Leftrightarrow {\rm{2sin(\alpha ) = }} - 2\]

\[ \Leftrightarrow {\rm{sin(\alpha ) = }} - 1\]

Đáp án cần chọn là: A