\(-1\le sin\left(x+\dfrac{\pi}{3}\right)\le1\Rightarrow-2\le2sin\left(x+\dfrac{\pi}{3}\right)\le2\)
\(\Rightarrow1\le y\le5\)
\(y_{min}=1\) khi \(sin\left(x+\dfrac{\pi}{3}\right)=1\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)
\(y_{max}=5\) khi \(sin\left(x+\dfrac{\pi}{3}\right)=-1\Rightarrow x=-\dfrac{5\pi}{6}+k2\pi\)
Lời giải:
Vì $\sin (x+\frac{\pi}{3})\in [-1;1]$
$\Rightarrow y=-2\sin (x+\frac{\pi}{3})+3\in [1;5]$
Vậy $y_{\min}=1$ và $y_{\max}=5$
Ta có \(-1\le\sin\left(x-\dfrac{\pi}{3}\right)\le1\Leftrightarrow1\le-2\sin\left(x-\dfrac{\pi}{3}\right)-3\le5\)
Vậy \(y_{min}=1\) khi \(\sin\left(x-\dfrac{\pi}{3}\right)=1\)
\(y_{max}=5\) khi \(\sin\left(x-\dfrac{\pi}{3}\right)=-1\)
