Đáp án + Giải thích các bước giải:

$\textbf{f}\bigg) y = \cos 2x + 5 \sin x + 2$

$= 1 - 2\sin^2 x + 5 \sin x + 2$

$= -2\sin^2 x + 5 \sin x + 3$

$= -\dfrac{1}{2}\bigg(4 \sin^2 x - 10\sin x - 6\bigg)$

$= -\dfrac{1}{2}\bigg[4\sin^2 x - 10\sin x + \dfrac{25}{4} - \dfrac{49}{4}\bigg)$

$= -\dfrac{1}{2}\bigg(2 \sin x - \dfrac{5}{2}\bigg)^2 + \dfrac{49}{8}$

Ta có: $1 \le 2 \sin x \le 2$ với mọi $x \in \bigg[\dfrac{\pi}{3}; \dfrac{5\pi}{6}\bigg]$

$\Leftrightarrow -\dfrac{3}{2} \le 2 \sin x - \dfrac{5}{2} \le -\dfrac{1}{2}$ với mọi $x \in \bigg[\dfrac{\pi}{3}; \dfrac{5\pi}{6}\bigg]$

$\Leftrightarrow \dfrac{1}{4} \le \bigg(2\sin x - \dfrac{5}{2}\bigg)^2 \le \dfrac{9}{4}$ với mọi $x \in \bigg[\dfrac{\pi}{3}; \dfrac{5\pi}{6}\bigg]$

$\Leftrightarrow -\dfrac{9}{8} \le -\dfrac{1}{2}\bigg(2 \sin x - \dfrac{5}{2}\bigg)^2 \le -\dfrac{1}{8}$ với mọi $x \in \bigg[\dfrac{\pi}{3}; \dfrac{5\pi}{6}\bigg]$

$\Leftrightarrow 5 \le -\dfrac{1}{2}\bigg(2\sin x - \dfrac{5}{2}\bigg)^2 + \dfrac{49}{8} \le 6$ với mọi $x \in \bigg[\dfrac{\pi}{3}; \dfrac{5\pi}{6}\bigg]$

$\Leftrightarrow 5 \le y \le 6$ với mọi $x \in \bigg[\dfrac{\pi}{3}; \dfrac{5\pi}{6}\bigg]$

$y_{\rm max} = 6$ khi $\sin x = 1 \Leftrightarrow x = \dfrac{\pi}{2} + k2\pi (k \in \mathbb{Z})$

$y_{\rm min} = 5$ khi $\sin x = \dfrac{1}{2} \Leftrightarrow \left[ \begin{array}{l}x=\dfrac{\pi}{6} + k2\pi\\x=\dfrac{5\pi}{6} + k2\pi \end{array} \right. (k \in \mathbb{Z})$

Xét $x = \dfrac{\pi}{2} + k2\pi \in \bigg[\dfrac{\pi}{3}; \dfrac{5\pi}{6}\bigg]$, ta có:

$\dfrac{\pi}{3} \le \dfrac{\pi}{2} + k2\pi \le \dfrac{5\pi}{6}$

$\Leftrightarrow \dfrac{1}{6} \le \dfrac{1}{4} + k \le \dfrac{5}{12}$

$\Leftrightarrow -\dfrac{1}{12} \le k \le \dfrac{1}{6}$

Vì $k \in \mathbb{Z} \Rightarrow k = 0$

Xét $x = \dfrac{\pi}{6} + k2\pi \in \bigg[\dfrac{\pi}{3}; \dfrac{5\pi}{6}\bigg]$, ta có:

$\dfrac{\pi}{3} \le \dfrac{\pi}{6} + k2\pi \le \dfrac{5\pi}{6}$

$\Leftrightarrow \dfrac{1}{6} \le \dfrac{1}{12} + k \le \dfrac{5}{12}$

$\Leftrightarrow -\dfrac{1}{12} \le k \le \dfrac{1}{3}$

Vì $k \in \mathbb{Z} \Rightarrow$ Không có $k$ thoả mãn

Xét $x = \dfrac{5\pi}{6} + k2\pi \in \bigg[\dfrac{\pi}{3}; \dfrac{5\pi}{6}\bigg]$, ta có:

$\dfrac{\pi}{3} \le \dfrac{5\pi}{6} + k2\pi \le \dfrac{5\pi}{6}$

$\Leftrightarrow \dfrac{1}{6} \le \dfrac{5}{12} + k \le \dfrac{5}{12}$

$\Leftrightarrow -\dfrac{1}{4} \le k \le 0$

Vì $k \in \mathbb{Z} \Rightarrow k = 0$

Vậy $y_{\rm \min} = 5$ khi $x = \dfrac{5\pi}{6}$ và $y_{\rm max} = 6$ khi $x = \dfrac{\pi}{2}$