Đáp án đúng: A
Giải chi tiết:Xét phương trình \(\cos 2x.\sin 5x + 1 = 0 \Leftrightarrow \cos 2x.\sin 5x = - 1\) (1), ta có:
\(0 \le \left| {\cos 2x} \right| \le 1,\,\,0 \le \left| {\sin 5x} \right| \le 1,\,\,\forall x \Rightarrow 0 \le \left| {\cos 2x.\sin 5x} \right| \le 1,\,\,\forall x \Leftrightarrow - 1 \le \cos 2x.\sin 5x \le 1,\,\,\forall x\)
\(\cos 2x.\sin 5x = - 1 \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}\cos 2x = 1\\\sin 5x = - 1\end{array} \right.\\\left\{ \begin{array}{l}\cos 2x = - 1\\\sin 5x = 1\end{array} \right.\end{array} \right.\)
TH1: \(\left\{ \begin{array}{l}\cos 2x = 1\\\sin 5x = - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ {x = k\pi ,\,\,k \in Z} \right.\\\sin 5x = - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = k2\pi ,\,\,k \in Z\\x = \pi + k2\pi ,\,\,k \in Z\end{array} \right.\\\sin 5x = - 1\end{array} \right. \Rightarrow x \in \emptyset \)
TH2: \(\left\{ \begin{array}{l}\cos 2x = - 1\\\sin 5x = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{\pi }{2} + k\pi ,\,\,k \in Z\\\sin 5x = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{2} + k2\pi ,\,\,k \in Z\\x = - \frac{\pi }{2} + k2\pi ,\,\,k \in Z\end{array} \right.\\\sin 5x = 1\end{array} \right. \Leftrightarrow x = \frac{\pi }{2} + k2\pi ,\,\,k \in Z\)
Mà \(x \in \left[ { - \frac{\pi }{2};2\pi } \right] \Rightarrow - \frac{\pi }{2} \le \frac{\pi }{2} + k2\pi \le 2\pi ,\,\,k \in Z \Leftrightarrow - \frac{1}{2} \le k \le \frac{3}{4},\,\,k \in Z \Leftrightarrow k = 0\). Có 1 giá trị của k thỏa mãn.
Vậy, phương trình đã cho có 1 nghiệm trên đoạn \(\left[ { - \frac{\pi }{2};2\pi } \right]\).
Chọn: A
