Ta có: \(-1<=\sin\left(2x-\frac{\pi}{3}\right)\le1\)
=>\(-4<=4\cdot\sin\left(2x-\frac{\pi}{3}\right)\le4\)
=>\(-4+5<=4\cdot\sin\left(2x-\frac{\pi}{3}\right)+5\le4+5\)
=>\(1<=4\cdot\sin\left(2x-\frac{\pi}{3}\right)+5\le9\)
=>\(1<=\sqrt{4\cdot\sin\left(2x-\frac{\pi}{3}\right)}+5\le\sqrt9\)
=>\(1<=\sqrt{4\cdot\sin\left(2x-\frac{\pi}{3}\right)}+5\le3\)
=>1<=y<=3
Do đó, ta có:
\(y_{\min}\) =1 khi \(\sin\left(2x-\frac{\pi}{3}\right)=-1\)
=>\(2x-\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)
=>\(2x=-\frac{\pi}{2}+\frac{\pi}{3}+k2\pi=-\frac{\pi}{6}+k2\pi\)
=>\(x=-\frac{\pi}{12}+k\pi\)
\(y_{\max}=3\) khi \(\sin\left(2x-\frac{\pi}{3}\right)=1\)
=>\(2x-\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)
=>\(2x=\frac{\pi}{2}+\frac{\pi}{3}+k2\pi=\frac{5\pi}{6}+k2\pi\)
=>\(x=\frac{5\pi}{12}+k\pi\)
