Đáp án:
\(\begin{array}{l}
\max y = 1 \Leftrightarrow x = 0\\
\min y = - \dfrac{1}{2} \Leftrightarrow x = - \dfrac{\pi }{3}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
x \in \left[ { - \dfrac{\pi }{3};\dfrac{\pi }{6}} \right] \Rightarrow 2x \in \left[ { - \dfrac{{2\pi }}{3};\dfrac{\pi }{3}} \right]\\
\Rightarrow y = \cos 2x \in \left[ { - \dfrac{1}{2};1} \right]\\
\Rightarrow \max y = 1 \Leftrightarrow \cos 2x = 1 \Leftrightarrow 2x = k2\pi \Leftrightarrow x = k\pi \,\,\left( {k \in Z} \right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \in \left[ { - \dfrac{\pi }{3};\dfrac{\pi }{6}} \right] \Leftrightarrow x = 0\\
\,\,\,\,\,\,\min y = - \dfrac{1}{2} \Leftrightarrow \cos 2x = - \dfrac{1}{2} \Leftrightarrow 2x = \pm \dfrac{{2\pi }}{3} + k2\pi \Leftrightarrow x = \pm \dfrac{\pi }{3} + k\pi \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \in \left[ { - \dfrac{\pi }{3};\dfrac{\pi }{6}} \right] \Leftrightarrow x = - \dfrac{\pi }{3}
\end{array}\)
