Đáp án:
\[\left\{ \begin{array}{l}
{y_{\min }} = - 2 \Leftrightarrow x = \frac{{ - \pi }}{6} + k2\pi \\
{y_{\max }} = 2 \Leftrightarrow x = \frac{{5\pi }}{6} + k2\pi
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \sin x - \sqrt 3 \cos x\\
= 2.\left( {\frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x} \right)\\
= 2.\left( {\sin x.\cos \frac{\pi }{3} - \sin \frac{\pi }{3}.\cos x} \right)\\
= 2.\sin \left( {x - \frac{\pi }{3}} \right)\\
- 1 \le \sin \left( {x - \frac{\pi }{3}} \right) \le 1 \Rightarrow - 2 \le 2\sin \left( {x - \frac{\pi }{3}} \right) \le 2\\
\Rightarrow - 2 \le y \le 2\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = - 2 \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = - 1 \Leftrightarrow x - \frac{\pi }{3} = \frac{{ - \pi }}{2} + k2\pi \Leftrightarrow x = \frac{{ - \pi }}{6} + k2\pi \\
{y_{\max }} = 2 \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 1 \Leftrightarrow x - \frac{\pi }{3} = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{{5\pi }}{6} + k2\pi
\end{array} \right.
\end{array}\)
Vậy \(\left\{ \begin{array}{l}
{y_{\min }} = - 2 \Leftrightarrow x = \frac{{ - \pi }}{6} + k2\pi \\
{y_{\max }} = 2 \Leftrightarrow x = \frac{{5\pi }}{6} + k2\pi
\end{array} \right.\)
