ta có
\(\begin{array}{l} \left\{ {\begin{array}{*{20}{c}} {{{\sin }^2}x + {{\cos }^2}x = 1}\\ {\cos x = 2\sin x} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {{{\sin }^2}x + 4{{\sin }^2}x = 1}\\ {\cos x = 2\sin x} \end{array}} \right.\\ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {{{\sin }^2}x = \frac{1}{5}}\\ {\cos x = 2\sin x} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {\left[ {\begin{array}{*{20}{c}} {\sin x = \frac{1}{{\sqrt 5 }}}\\ {\sin x = - \frac{1}{{\sqrt 5 }}} \end{array}} \right.}\\ {\cos x = 2\sin x} \end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}} {\sin x = \frac{1}{{\sqrt 5 }} \Rightarrow \cos x = \frac{2}{{\sqrt 5 }}}\\ {\sin x = - \frac{1}{{\sqrt 5 }} \Rightarrow {\mathop{\rm cosx}olimits} = - \frac{2}{{\sqrt 5 }}} \end{array}} \right. \end{array}\)
Vậy \(\sin x.cosx = \frac{2}{5}\)
