\(y=2-2.\left(2sinx.cosx\right)=2-2sin2x\)

Do \(-1\le sin2x\le1\Rightarrow0\le y\le4\)

\(y_{min}=0\) khi \(sin2x=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

\(y_{max}=4\) khi \(sin2x=-1\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)

\(y=2\left(1-cos2x\right)-cos2x=2-3cos2x\)

Do \(-1\le cos2x\le1\Rightarrow-1\le y\le5\)

\(y_{min}=-1\) khi \(cos2x=1\Leftrightarrow x=k\pi\)

\(y_{max}=5\) khi \(cos2x=-1\Rightarrow x=\dfrac{\pi}{2}+k\pi\)

1.

\(sin\left(x-\dfrac{\pi}{3}\right)\in\left[-1;1\right]\)

\(\Rightarrow y=2sin\left(x-\dfrac{\pi}{3}\right)+3\in\left[1;5\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=1\\y_{max}=5\end{matrix}\right.\)

2.

\(cos2x\in\left[-1;1\right]\)

\(\Rightarrow y=3-\dfrac{1}{2}cos2x\in\left[\dfrac{5}{2};\dfrac{7}{2}\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=\dfrac{5}{2}\\y_{max}=\dfrac{7}{2}\end{matrix}\right.\)

3.

\(y=\dfrac{1+3cos^2x}{2}=\dfrac{\dfrac{3}{2}\left(2cos^2x-1\right)+\dfrac{5}{2}}{2}=\dfrac{3}{4}cos2x+\dfrac{5}{4}\)

Vì \(cos2x\in\left[-1;1\right]\Rightarrow y=\dfrac{3}{4}cos2x+\dfrac{5}{4}\in\left[\dfrac{1}{2};2\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=\dfrac{1}{2}\\y_{max}=2\end{matrix}\right.\)

Do \(0\le cos^2x\le1\Rightarrow\dfrac{1}{2}\le y\le2\)

\(y_{min}=\dfrac{1}{2}\) khi \(cosx=0\Rightarrow x=\dfrac{\pi}{2}+k\pi\)

\(y_{max}=2\) khi \(cos^2x=1\Rightarrow sinx=0\Rightarrow x=k\pi\)

\(y=4\left(1-sin^2x\right)+2sinx+2=-4sin^2x+2sinx+6\)

Đặt \(sinx=t\in\left[-1;1\right]\Rightarrow y=f\left(t\right)=-4t^2+2t+6\)

\(-\dfrac{b}{2a}=\dfrac{1}{4}\in\left[-1;1\right]\)

\(f\left(-1\right)=0\) ; \(f\left(\dfrac{1}{4}\right)=\dfrac{25}{4}\); \(f\left(1\right)=4\)

\(\Rightarrow y_{max}=\dfrac{25}{4}\) khi \(sinx=\dfrac{1}{4}\)

\(y_{min}=0\) khi \(sinx=-1\)

Ta có: \(y=4cos^2x+2sinx+2=4-4sin^2x+2sinx+2=-4sin^2x+2sinx+6=-\left(4sin^2x-2sinx+\dfrac{1}{16}-\dfrac{1}{16}-6\right)=-\left(2sin^2x-\dfrac{1}{4}\right)^2+\dfrac{97}{16}\)

Ta có: \(-\left(2sin^2x-\dfrac{1}{4}\right)^2\le0\Rightarrow y\le\dfrac{97}{16}\)

Vậy \(y_{max}=\dfrac{97}{16}\)

\(\Leftrightarrow\left(sinx-3\right)\left(2sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=3>1\left(loại\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)