Đáp án:
$min_y=-\dfrac{9}{4} \Leftrightarrow \left[\begin{array}{l} x=\arcsin \left(-\dfrac{1}{2\sqrt{2}} \right)-\dfrac{\pi}{4}+ k 2 \pi(k \in \mathbb{Z})\\x=\dfrac{3\pi}{4}-\arcsin \left(-\dfrac{1}{2\sqrt{2}} \right)+ k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\ max_y=\sqrt{2} \Leftrightarrow x=\dfrac{\pi}{4}+ k 2 \pi(k \in \mathbb{Z}).$
Giải thích các bước giải:
$y= \sin x+\cos x+2\sin x\cos x-1\\ = \sin x+\cos x+2\sin x\cos x+1-2\\ =\sin x+\cos x+\sin^2x+2\sin x\cos x+\cos^2x-2\\ =\sin x+\cos x+(\sin x+\cos x)^2-2\\ =(\sin x+\cos x)^2+\sin x+\cos x-2\\ =2\sin \left(x+\dfrac{\pi}{4}\right)^2+\sqrt{2} \sin \left(x+\dfrac{\pi}{4}\right)-2\\ t=\sin \left(x+\dfrac{\pi}{4}\right); t\in[-1;1]\\ y=2t^2+\sqrt{2}t-2\\ y'=4t+\sqrt{2}\\ y'=0 \Leftrightarrow t=-\dfrac{1}{2\sqrt{2}}\\ BBT$
\begin{array}{|c|ccccccccc|} \hline x&-1&&-\dfrac{1}{2\sqrt{2}}&&1\\\hline y'&&-&0&+&\\\hline &-\sqrt{2}&&&&\sqrt{2}\\y&&\searrow&&\nearrow&\\&&&-\dfrac{9}{4}\\\hline\end{array}
Dựa vào BBT,
$min_y=-\dfrac{9}{4} \Leftrightarrow t=-\dfrac{1}{2\sqrt{2}} \Leftrightarrow \sin \left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{2\sqrt{2}} \Leftrightarrow \left[\begin{array}{l} x+\dfrac{\pi}{4}=\arcsin \left(-\dfrac{1}{2\sqrt{2}} \right)+ k 2 \pi(k \in \mathbb{Z})\\x+\dfrac{\pi}{4}=\pi-\arcsin \left(-\dfrac{1}{2\sqrt{2}} \right)+ k 2 \pi(k \in \mathbb{Z})\end{array} \right.\Leftrightarrow \left[\begin{array}{l} x=\arcsin \left(-\dfrac{1}{2\sqrt{2}} \right)-\dfrac{\pi}{4}+ k 2 \pi(k \in \mathbb{Z})\\x=\dfrac{3\pi}{4}-\arcsin \left(-\dfrac{1}{2\sqrt{2}} \right)+ k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\ max_y=\sqrt{2} \Leftrightarrow t=1\Leftrightarrow \sin \left(x+\dfrac{\pi}{4}\right)=1 \Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+ k 2 \pi(k \in \mathbb{Z}) \Leftrightarrow x=\dfrac{\pi}{4}+ k 2 \pi(k \in \mathbb{Z}).$
