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Đặt \(sin^24x=t\left(t\in\left[0;1\right]\right)\)
\(y=1-8sin^22x.cos^22x+2sin^42x\)
\(=1-2sin^24x+2sin^42x\)
\(\Rightarrow y=f\left(t\right)=1-2t+2t^2\)
\(y_{min}=min\left\{f\left(0\right);f\left(1\right);f\left(\dfrac{1}{2}\right)\right\}=\dfrac{1}{2}\)
\(y_{max}=max\left\{f\left(0\right);f\left(1\right);f\left(\dfrac{1}{2}\right)\right\}=1\)
\(y=2cos^2x-2\sqrt{3}sinx.cosx+1\)
\(=2cos^2x-1-2\sqrt{3}sinx.cosx+2\)
\(=cos2x-\sqrt{3}sin2x+2\)
\(=2\left(\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x\right)+2\)
\(=2cos\left(2x+\dfrac{\pi}{3}\right)+2\)
Ta có: \(cos\left(2x+\dfrac{\pi}{3}\right)\in\left[-1;1\right]\)
\(\Rightarrow min=0\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=-1\Leftrightarrow2x+\dfrac{\pi}{3}=\pi+k2\pi\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)
\(\Rightarrow max=4\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=1\Leftrightarrow2x+\dfrac{\pi}{3}=k2\pi\Leftrightarrow x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)
\(y=2cos^2x-\sqrt{3}sin2x+1=cos2x-\sqrt{3}sin2x+2\)
\(y=2.cos\left(2x+\dfrac{\pi}{3}\right)+2\)
\(\forall x\in R->-1\le cos\left(2x+\dfrac{\pi}{3}\right)\)
=> \(Min_y=2.\left(-1\right)+2=0\)
Mặt khác, theo Bunhiacopxki:
\(\left(cos2x+\sqrt{3}sin2x\right)^2\le\left(1^2+\sqrt{3}^2\right)\left(cos^22x+sin^22x\right)=4\)
=>\(Max_y=4\)
