Đáp án:
$a) \, \begin{cases}miny = - 1\Leftrightarrow x = \dfrac{2\pi}{3} + k2\pi\\maxy = 1\Leftrightarrow x = - \dfrac{\pi}{3} + k2\pi\end{cases} \quad (k \in \Bbb Z)$
$b) \, \begin{cases}miny = -2\Leftrightarrow x = -\dfrac{\pi}{3} + k2\pi\\maxy = 2 \Leftrightarrow x = \dfrac{2\pi}{3} + k2\pi\end{cases}\quad (k \in \Bbb Z)$
$c)\, \begin{cases}miny = 2 \Leftrightarrow x = \dfrac{\pi}{2} + k2\pi\\maxy = 6 \Leftrightarrow x = - \dfrac{\pi}{2} + k2\pi\end{cases}\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}a)\,y = \cos\left(x + \dfrac{2\pi}{3}\right) + \cos x\\ =\cos x\cos\dfrac{2\pi}{3} - \sin x\sin\dfrac{2\pi}{3} + \cos x\\ = -\dfrac{1}{2}\cos x-\dfrac{\sqrt3}{2}\sin x + \cos x\\ = \dfrac{1}{2}\cos x - \dfrac{\sqrt3}{2}\sin x\\ = \cos\dfrac{\pi}{3}\cos x - \sin\dfrac{\pi}{3}\sin x\\ = \cos\left(x + \dfrac{\pi}{3}\right)\\ Ta\,\,có: -1 \leq\cos\left(x + \dfrac{\pi}{3}\right) \leq 1\\ Hay \,\,-1 \leq y \leq 1\\ Vậy\,\,miny = - 1 \Leftrightarrow \cos\left(x + \dfrac{\pi}{3}\right) = -1 \Leftrightarrow x = \dfrac{2\pi}{3} + k2\pi\\ maxy = 1 \Leftrightarrow \cos\left(x + \dfrac{\pi}{3}\right) = 1 \Leftrightarrow x = - \dfrac{\pi}{3} + k2\pi \quad (k \in \Bbb Z)\\ b)\, y = \sqrt3\sin x - \cos x\\ = 2\left(\dfrac{\sqrt3}{2}\sin x - \dfrac{1}{2}\cos x\right)\\ = 2\sin\left(x - \dfrac{\pi}{6}\right)\\ Ta\,\,có:\\ -1 \leq \sin\left(x - \dfrac{\pi}{6}\right) \leq 1\\ \Leftrightarrow -2 \leq 2\sin\left(x - \dfrac{\pi}{6}\right) \leq 2\\ Hay\,\,-2 \leq y \leq 2 \\Vậy\,\,miny = -2 \Leftrightarrow \sin\left(x - \dfrac{\pi}{6}\right) = -1 \Leftrightarrow x = -\dfrac{\pi}{3} + k2\pi\\ maxy = 2 \Leftrightarrow \sin\left(x - \dfrac{\pi}{6}\right) = 1 \Leftrightarrow x = \dfrac{2\pi}{3} + k2\pi \quad (k \in \Bbb Z)\\ c)\,y = \sin^2x + 2\cos^2x+ 2\sin x + 3\\ = \sin^2x + 2( 1 - \sin^2x) + 2\sin x + 3\\ = -\sin^2x + 2\sin x + 5\\ = -(\sin^2x + 2\sin x + 1) + 6\\ = - (\sin x + 1)^2 + 6\\ Ta\,\,có:\\ -1 \leq \sin x \leq 1\\ \Leftrightarrow 0 \leq \sin x + 1 \leq 2\\ \Leftrightarrow 0 \leq (\sin x + 1)^2 \leq 4\\ \Leftrightarrow -4 \leq -(\sin x + 1)^2 \leq 0\\ \Leftrightarrow 2 \leq - (\sin x + 1)^2 + 6 \leq 6\\ Hay\,\,2 \leq y \leq 6\\ Vậy\,\,miny = 2 \Leftrightarrow \sin x = 1 \Leftrightarrow x = \dfrac{\pi}{2} + k2\pi\\ maxy = 6 \Leftrightarrow \sin x = - 1 \Leftrightarrow x = - \dfrac{\pi}{2} + k2\pi\quad (k \in \Bbb Z)\end{array}$
