Ta có $-1\le \sin x\le 1$
Điều kiện xác định:
$\begin{array}{l} \left\{ \begin{array}{l} \cos 4x \ge 1\\ \sin x \ne 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \cos 4x = 1\\ \sin x \ne 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 4x = \dfrac{\pi }{2} + k2\pi \\ x \ne \dfrac{\pi }{2} + k2\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\ x \ne \dfrac{\pi }{2} + k2\pi \end{array} \right.\left( {do\,\sin x \le 1} \right)\\ D = \left\{ {\dfrac{\pi }{8} + \dfrac{{k\pi }}{2}|k \in Z} \right\} \end{array}$
Do $\sin x\le 1$ nên $4-3sinx \ge 1>0$
$\begin{array}{l}
\Rightarrow y = 3 + \left| {4 - 3\sin x} \right| = 3 + 4 - 3\sin x = 7 - 3\sin x\\
- 1 \le \sin x \le 1 \Rightarrow - 3 \le 3\sin x \le 3\\
\Rightarrow 4 \le y \le 10 \Rightarrow \max y = 10,\min y = 4
\end{array}$
