Phương pháp giải:

Biến đổi:

\(\begin{array}{l}
{\sin ^6}x + {\cos ^6}x = \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x + {{\cos }^4}x - {{\sin }^2}x{{\cos }^2}x} \right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 3{\sin ^2}x{\cos ^2}x\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \frac{3}{4}{\sin ^2}2x\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \frac{3}{4} \cdot \frac{{1 - \cos 4x}}{2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{5 + 3\cos 4x}}{8}
\end{array}\)

Lời giải chi tiết:

Biến đổi:

\(\begin{array}{l}y = {\sin ^6}x + {\cos ^6}x + 7\sin 4x\\\,\,\,\,\, = \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x + {{\cos }^4}x - {{\sin }^2}x{{\cos }^2}x} \right) + 7\sin 4x\\\,\,\,\,\, = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 3{\sin ^2}x{\cos ^2}x + 7\sin 4x\\\,\,\,\,\, = 1 - \frac{3}{4}{\sin ^2}2x + 7\sin 4x\\\,\,\,\,\,\, = 1 - \frac{3}{4} \cdot \frac{{1 - \cos 4x}}{2} + 7\sin 4x\\\,\,\,\,\,\, = \frac{5}{8} + 7\sin 4x + \frac{3}{8}\cos 4x\end{array}\)

Theo bất đẳng thức Bunhiacopxki:

\(\begin{array}{l}\;\;\;\;{\left( {7\sin 4x + \frac{3}{4}\cos 4x} \right)^2} \le \left( {{{\sin }^2}4x + {{\cos }^2}4x} \right)\left( {{7^2} + {{\left( {\frac{3}{8}} \right)}^2}} \right)\\ \Leftrightarrow \;\;\;{\left( {7\sin 4x + \frac{3}{8}\cos 4x} \right)^2} \le \frac{{3145}}{{64}}\\ \Leftrightarrow \left| {7\sin 4x + \frac{3}{8}\cos 4x} \right| \le \frac{{\sqrt {3145} }}{8}\\ \Leftrightarrow  - \frac{{\sqrt {3145} }}{8} \le 7\sin 4x + \frac{3}{8}\cos 4x \le \frac{{\sqrt {3145} }}{8}\\ \Leftrightarrow \frac{5}{8} - \frac{{\sqrt {3145} }}{8} \le 7\sin 4x + \frac{3}{8}\cos 4x + \frac{5}{8} \le \frac{{\sqrt {3145} }}{8} + \frac{5}{8}\\ \Leftrightarrow  - 6,38 \le y \le 7,63\end{array}\)

Các giá trị nguyên của \(y:\) \(y \in \left\{ { - 6; - 5;...; - 1;0;1;...7} \right\},\) có \(14\) giá trị.

Chọn A