Câu hỏi 19

\(\sin \left( {\cos 2x} \right) = 0\,\,\,\left( * \right) \Leftrightarrow \cos 2x = k\pi \,\,\left( {k \in \mathbb{Z}} \right)\,\,\,\left( 1 \right)\)

Do \( - 1 \le \cos 2x \le 1 \Leftrightarrow  - 1 \le k\pi  \le 1 \Leftrightarrow  - \dfrac{1}{\pi } \le k \le \dfrac{1}{\pi }\,\,\left( {k \in \mathbb{Z}} \right) \Leftrightarrow k = 0\)

\(\begin{array}{l} \Rightarrow \left( 1 \right) \Leftrightarrow \cos 2x = 0 \Leftrightarrow 2x = \dfrac{\pi }{2} + m\pi  \Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{m\pi }}{2}\,\,\,\,\left( {m \in \mathbb{Z}} \right)\\Do\,\,x \in \left[ {0;\,2\pi } \right] \Rightarrow 0 \le \dfrac{\pi }{4} + \dfrac{{m\pi }}{2} \le 2\pi  \Leftrightarrow  - \dfrac{1}{2} \le m \le \dfrac{7}{2} \Rightarrow m \in \left\{ {0;\,1;\,2;\,3} \right\}.\end{array}\)

Vậy phương trình đã cho có 4 nghiệm thỏa mãn bài toán.

Chọn A.