\(y=1-cos2x+2sin2x+6=2sin2x-cos2x+7\)

\(y=\sqrt{5}\left(\dfrac{2}{\sqrt{5}}sin2x-\dfrac{1}{\sqrt{5}}cos2x\right)+7\)

Đặt \(\dfrac{2}{\sqrt{5}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)

\(y=\sqrt{5}sin\left(2x-a\right)+7\)

\(\Rightarrow-\sqrt{5}+7\le y\le\sqrt{5}+7\)

Đặt \(sinx=t\in\left[-1;1\right]\)

\(\Rightarrow y=f\left(t\right)=4t^2-3t-1\)

Xét hàm \(f\left(t\right)\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=\dfrac{3}{8}\in\left[-1;1\right]\)

\(f\left(-1\right)=6\) ; \(f\left(\dfrac{3}{8}\right)=-\dfrac{25}{16}\) ; \(f\left(1\right)=0\)

\(\Rightarrow y_{min}=-\dfrac{25}{16}\) khi \(sinx=\dfrac{3}{8}\)

\(y_{max}=6\) khi \(sinx=-1\)

Đặt \(sinx=t\left(t\in\left[-1;1\right]\right)\).

\(\Rightarrow y=f\left(t\right)=-2t^2+3t-1\)

\(\Rightarrow y_{min}=min\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{3}{4}\right)\right\}=f\left(-1\right)=-6\)

\(y_{max}=max\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{3}{4}\right)\right\}=f\left(\dfrac{3}{4}\right)=\dfrac{1}{8}\)

a, \(y=2sin^2x-cos2x=1-2cos2x\)

Vì \(cos2x\in\left[-1;1\right]\Rightarrow y=2sin^2x-cos2x\in\left[-1;3\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=-1\\y_{max}=3\end{matrix}\right.\)

a, \(y=3-4sin^2x.cos^2x=3-sin^22x\)

Đặt \(sin2x=t\left(t\in\left[-1;1\right]\right)\).

\(\Rightarrow y=f\left(t\right)=3-t^2\)

\(\Rightarrow y_{min}=minf\left(t\right)=2\)

\(y_{max}=maxf\left(t\right)=3\)

b, \(y=f\left(t\right)=\dfrac{-2}{3t-5}\left(t\in\left[0;1\right]\right)\)

\(\Rightarrow y_{min}=minf\left(t\right)=\dfrac{2}{5}\)

\(y_{max}=maxf\left(t\right)=1\)

\(y=\sin^4x+\cos^4x\\ =1-2\sin^2x\cdot\cos^2x\\ =1-\dfrac{1}{2}\sin^22x\\ 0\le\sin^22x\le1\\ \Leftrightarrow\dfrac{1}{2}\le y\le1\\ y_{min}=\dfrac{1}{2}\Leftrightarrow\sin^22x=1\Leftrightarrow x=\dfrac{k\pi}{2}\pm\dfrac{\pi}{4}\\ y_{max}=1\Leftrightarrow\sin^22x=0\Leftrightarrow x=k\pi\)

\(y=3\sin x+4\cos x\\ =5\left(\dfrac{3\sin x}{5}+\dfrac{4\cos x}{5}\right)\\ =5\cos\left(x-a\right),\forall\cos a=\dfrac{4}{5},\sin a=\dfrac{3}{5}\\ -1\le\cos\left(x-a\right)\le1\\ \Leftrightarrow-5\le y\le5\\ y_{min}=-5\Leftrightarrow\cos\left(x-a\right)=-1\\ y_{max}=5\Leftrightarrow\cos\left(x-a\right)=1\)

\(y=sin^4x+cos^4x\)

Ta có: \(0\le sin^4x\le1\)

\(0\le cos^4x\le1\)

\(0\le sin^4x+cos^4x\le2\)

Vây GTNN là 0, GTLN là 2

y=3sinx+4cosx

\(-3\le3sinx\le3\\ -4\le4cosx\le4\\ -7\le3sinx+4cosx\le7\)

Vậy GTNN là -7, GTLN là 7