Đáp án:
$\begin{cases}miny = 2 \Leftrightarrow \left[\begin{array}{l}x = \arcsin\left(-\dfrac{4}{5}\right) + k2\pi\\x = \pi - \arcsin\left(-\dfrac{4}{5}\right) + k2\pi \end{array}\right.\\maxy = 5 \Leftrightarrow x = \dfrac{\pi}{2} + k2\pi\end{cases}\,\,\,\,\,\,(k \in \Bbb Z)$
Giải thích các bước giải:
$y = 2 + \sqrt{5\sin x + 4}$
Ta có:
$+) \,\,\sqrt{5\sin x + 4} \geq 0$ (theo điều kiện để căn thức có nghĩa)
$\Leftrightarrow 2 + \sqrt{5\sin x + 4} \geq 2$
Hay $y \geq 2$
Dấu = xảy ra $\Leftrightarrow \sqrt{5\sin x + 4} = 0$
$\Leftrightarrow \sin x = -\dfrac{4}{5}$
$\Leftrightarrow \left[\begin{array}{l}x = \arcsin\left(-\dfrac{4}{5}\right) + k2\pi\\x = \pi - \arcsin\left(-\dfrac{4}{5}\right) + k2\pi \end{array}\right.\,\,\,\,\,(k \in \Bbb Z)$
$+)\,\, \sin x \leq 1$
$\Leftrightarrow 5\sin x + 4 \leq 9$
$\Leftrightarrow \sqrt{5\sin x + 4} \leq 3$
$\Leftrightarrow 2 + \sqrt{5\sin x + 4} \leq 5$
Hay $y \leq 5$
Dấu = xảy ra $\sin x = 1 \Leftrightarrow x = \dfrac{\pi}{2} + k2\pi\,\,\,\,(k \in \Bbb Z)$
Vậy $miny = 2 \Leftrightarrow \left[\begin{array}{l}x = \arcsin\left(-\dfrac{4}{5}\right) + k2\pi\\x = \pi - \arcsin\left(-\dfrac{4}{5}\right) + k2\pi \end{array}\right.\,\,\,\,\,(k \in \Bbb Z)$
$maxy = 5 \Leftrightarrow x = \dfrac{\pi}{2} + k2\pi\,\,\,\,(k \in \Bbb Z)$
