Đáp án:
\(\min = 1,\,\,\max = \sqrt 5 + 1\)
Giải thích các bước giải:
\(\begin{array}{l}
6\sin x - 8\cos x = 5\left( {\frac{3}{5}\sin x - \frac{4}{5}\cos x} \right)\\
Do\,\,{\left( {\frac{3}{5}} \right)^2} + {\left( {\frac{4}{5}} \right)^2} = 1 \Rightarrow Dat\,\left\{ \begin{array}{l}
\cos \alpha = \frac{3}{5}\\
\sin \alpha = \frac{4}{5}
\end{array} \right.\\
\Rightarrow 6\sin x - 8\cos x = 5\left( {\sin x\cos \alpha - \cos x\sin \alpha } \right) = 5\sin \left( {x - \alpha } \right)\\
\Rightarrow - 5 \le 6\sin x - 8\cos x \le 5\\
\Rightarrow 0 \le \sqrt {6\sin x - 8\cos x} \le \sqrt 5 \\
\Rightarrow 1 \le \sqrt {6\sin x - 8\cos x} + 1 \le \sqrt 5 + 1\\
\Rightarrow \min = 1,\,\,\max = \sqrt 5 + 1
\end{array}\)
