Đáp án:

$1)max_y=1 \Leftrightarrow x=\dfrac{\pi}{3} +k 2 \pi(k \in \mathbb{Z})\\ min_y=-2\Leftrightarrow x=\dfrac{4\pi}{3} +k 2 \pi(k \in \mathbb{Z})\\ 2)max_y= \sqrt{2} -3 \Leftrightarrow x=\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\\ min_y=-3 \Leftrightarrow x=-\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\\ 3)max_y= 3 \Leftrightarrow x=\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\\ min_y=-1 \Leftrightarrow x=-\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\\ 4)max_y= 2 \Leftrightarrow x=k 2 \pi(k \in \mathbb{Z})\\ min_y=-4 \Leftrightarrow  x=\pi+k 2 \pi(k \in \mathbb{Z})\\ 5)max_y= 10 \Leftrightarrow x=\pi+k 2 \pi(k \in \mathbb{Z})\\ min_y=1  \Leftrightarrow x=\pm \dfrac{\pi}{3}+k 2 \pi(k \in \mathbb{Z})\\ 8) max_y= \sqrt{5} \Leftrightarrow x=\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\\ min_y=1  \Leftrightarrow x=-\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z}).$

Giải thích các bước giải:

$1)y=2\cos \left(x-\dfrac{\pi}{3}\right)-1\\ -1 \le \cos \left(x-\dfrac{\pi}{3}\right) \le 1\\ \Rightarrow -2 \le \cos \left(x-\dfrac{\pi}{3}\right) \le 2\\ \Rightarrow -2 \le \cos \left(x-\dfrac{\pi}{3}\right) -1 \le 1\\ \Leftrightarrow  -2 \le y \le 1\\ \Rightarrow max_y=1 \Leftrightarrow \cos \left(x-\dfrac{\pi}{3}\right)=1 \Leftrightarrow x-\dfrac{\pi}{3} =k 2 \pi(k \in \mathbb{Z})\Leftrightarrow x=\dfrac{\pi}{3} +k 2 \pi(k \in \mathbb{Z})\\ min_y=-2 \Leftrightarrow \cos \left(x-\dfrac{\pi}{3}\right)=-1 \Leftrightarrow x-\dfrac{\pi}{3} =\pi+k 2 \pi(k \in \mathbb{Z})\Leftrightarrow x=\dfrac{4\pi}{3} +k 2 \pi(k \in \mathbb{Z})\\ 2)y=\sqrt{1+\sin x}-3\\ -1 \le \sin x \le 1\\ \Rightarrow 0 \le 1+\sin x \le 2\\ \Rightarrow 0 \le \sqrt{1+\sin x} \le \sqrt{2}\\ \Rightarrow -3 \le \sqrt{1+\sin x} -3 le \sqrt{2} -3\\ \Leftrightarrow -3 \le y le \sqrt{2} -3\\ \Rightarrow max_y= \sqrt{2} -3 \Leftrightarrow \sin x=1 \Leftrightarrow x=\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\\ min_y=-3 \Leftrightarrow \sin x=-1 \Leftrightarrow x=-\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\\ 3)y=2\sin x+1\\ -1 \le \sin x \le 1\\ \Rightarrow -2 \le 2\sin x \le 2\\ \Rightarrow -1 \le 2\sin x +1 \le 3\\ \Leftrightarrow -1 \le y \le 3\\ \Rightarrow max_y= 3 \Leftrightarrow \sin x=1 \Leftrightarrow x=\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\\ min_y=-1 \Leftrightarrow \sin x=-1 \Leftrightarrow x=-\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\\ 4)y=3\cos x-1\\ -1 \le \cos x \le 1\\ \Rightarrow -3 \le 3\cos x \le 3\\ \Rightarrow -4 \le 3\cos x-1 \le 2\\ \Leftrightarrow -4 \le y \le 2\\ \Rightarrow max_y= 2 \Leftrightarrow \cos x=1 \Leftrightarrow x=k 2 \pi(k \in \mathbb{Z})\\ min_y=-4 \Leftrightarrow \cos x=-1 \Leftrightarrow x=\pi+k 2 \pi(k \in \mathbb{Z})\\ 5)y=4\cos^2x-4\cos x+2\\ t=\cos x, t \in [-1;1]\\ y=4t^2-4t+2\\ y'=8t-4\\ y'=0 \Leftrightarrow t=\dfrac{1}{2}\\ BBT:$

\begin{array}{|c|ccccccccc|} \hline x&-1&&\dfrac{1}{2}&&1\\\hline y'&&-&0&+&\\\hline &10&&&&2\\y&&\searrow&&\nearrow&\\&&&1\\\hline\end{array}

$\Rightarrow max_y= 10 \Leftrightarrow t=\cos x=-1 \Leftrightarrow x=\pi+k 2 \pi(k \in \mathbb{Z})\\ min_y=1 \Leftrightarrow t=\cos x=\dfrac{1}{2} \Leftrightarrow x=\pm \dfrac{\pi}{3}+k 2 \pi(k \in \mathbb{Z})\\ 8)y=\sqrt{2\sin x+3}\\ -1 \le \sin x \le 1\\ \Rightarrow -2 \le 2\sin x \le 2\\ \Rightarrow 1 \le 2\sin x+3 \le 5\\ \Rightarrow 1 \le \sqrt{2\sin x+3} \le \sqrt{5}\\ \Leftrightarrow 1 \le y \le \sqrt{5}\\ \Rightarrow max_y= \sqrt{5} \Leftrightarrow \sin x=1 \Leftrightarrow x=\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\\ min_y=1 \Leftrightarrow \sin x=-1 \Leftrightarrow x=-\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z}).$