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\(y=\frac{2cos^2x+2sinx.cosx}{2+2sin^2x}=\frac{1+cos2x+sin2x}{3-cos2x}\)
\(\Rightarrow3y-y.cos2x=1+cos2x+sin2x\)
\(\Rightarrow sin2x+\left(y+1\right)cos2x=3y-1\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(1^2+\left(y+1\right)^2\ge\left(3y-1\right)^2\)
\(\Leftrightarrow8y^2-8y-1\le0\)
\(\Rightarrow\frac{2-\sqrt{6}}{4}\le y\le\frac{2+\sqrt{6}}{4}\)
\(0\le cos^2x\le1\Rightarrow2\le3-cos^2x\le3\)
\(\Rightarrow\frac{8}{3}\le y\le4\)
\(y_{min}=\frac{8}{3}\) khi \(cosx=0\)
\(y_{max}=4\) khi \(cos^2x=1\)
b/ \(0\le sin^23x\le1\Rightarrow1\le\sqrt{2-sin^23x}\le\sqrt{2}\)
\(\Rightarrow\frac{1}{\sqrt{2}}\le y\le1\)
\(y_{min}=\frac{1}{\sqrt{2}}\) khi \(sin3x=0\)
\(y_{max}=1\) khi \(sin^23x=1\)
c/ \(y=\sqrt{3}\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+sin2x+1\)
\(=-\sqrt{3}\left(cos^2x-sin^2x\right)+sin2x+1\)
\(=-\sqrt{3}cos2x+sin2x+1\)
\(=2\left(\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\right)+1=2sin\left(2x-\frac{\pi}{3}\right)+1\)
Do \(-1\le sin\left(2x-\frac{\pi}{3}\right)\le1\Rightarrow-1\le y\le3\)
\(y_{min}=-1\) khi \(sin\left(2x-\frac{\pi}{3}\right)=-1\)
\(y_{max}=3\) khi \(sin\left(2x-\frac{\pi}{3}\right)=1\)
a.
\(y=2sinx-\left(1-sin^2x\right)=sin^2x+2sinx-1=\left(sinx+1\right)^2-2\ge-2\)
\(\Rightarrow y_{min}=-2\)
\(y=sin^2x+2sinx-1=\left(sinx-1\right)\left(sinx+3\right)+2\le2\)
\(\Rightarrow y_{max}=2\)
b.
\(1\le3-2sinx\le5\Rightarrow6\le y\le5+\sqrt{5}\)
\(y_{min}=6\) ; \(y_{max}=5+\sqrt{5}\)
Ta có:
\(-1\le\sin2x\le1\)
=> \(\sqrt{4-2.\left(1\right)^5}-8\le\sqrt{4-2.\left(\sin2x\right)^5}-8\le\sqrt{4-2.\left(-1\right)^5}-8\)
=> \(\sqrt{2}-8\le\sqrt{4-2.\left(\sin2x\right)^5}-8\le\sqrt{6}-8\)
=> tìm ddc min và max
\(0\le cos^2\left(x-\frac{\pi}{4}\right)\le1\Rightarrow1\le y\le2\)
\(y_{min}=1\) khi \(cos\left(x-\frac{\pi}{4}\right)=0\)
\(y_{max}=2\) khi \(cos^2\left(x-\frac{\pi}{4}\right)=1\)
\(y=2cos\left(x+\frac{\pi}{6}\right)cos\left(\frac{\pi}{6}\right)=\sqrt{3}cos\left(x+\frac{\pi}{6}\right)\)
Do \(-1\le cos\left(x+\frac{\pi}{6}\right)\le1\) nên \(-\sqrt{3}\le y\le\sqrt{3}\)
\(y_{min}=-\sqrt{3}\) khi \(cos\left(x+\frac{\pi}{6}\right)=-1\)
\(y_{max}=\sqrt{3}\) khi \(cos\left(x+\frac{\pi}{6}\right)=1\)
a/ \(y=2\left(\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx\right)+5=2sin\left(x-\frac{\pi}{6}\right)+5\)
Do \(-1\le sin\left(x-\frac{\pi}{6}\right)\le1\Rightarrow3\le y\le7\)
b/ \(y=2cos\left(x+\frac{\pi}{6}\right)cos\left(-\frac{\pi}{6}\right)=\sqrt{3}cos\left(x+\frac{\pi}{6}\right)\)
Do \(-1\le cos\left(x+\frac{\pi}{6}\right)\le1\Rightarrow-\sqrt{3}\le y\le\sqrt{3}\)
c/ \(y=2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)+12=2sin\left(x+\frac{\pi}{3}\right)+12\)
Do \(-1\le sin\left(x+\frac{\pi}{3}\right)\le1\Rightarrow10\le y\le14\)
áp dụng bất đẳng thức Bunhia ta có :
\(\left(x+4y\right)^2\le\left(5^2+12^2\right)\left(\frac{x^2}{25}+\frac{y^2}{9}\right)=169\)
Vậy \(-13\le x+4y\le13\Rightarrow-8\le P\le18\)
vậy min bằng -8
max bằng 18
