\(y=2\left(\frac{1}{2}-\frac{1}{2}cos2x\right)+cos^22x=cos^22x-cos2x+1\)

\(=\left(cos2x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow y_{min}=\frac{3}{4}\) khi \(cos2x=\frac{1}{2}\)

\(y=cos^22x-2cos2x+cos2x-2+3\)

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

Do \(-1\le cos2x\le1\Rightarrow\left\{{}\begin{matrix}cos2x-2< 0\\cos2x+1\ge0\end{matrix}\right.\) \(\Rightarrow\left(cos2x-2\right)\left(cos2x+1\right)\le0\)

\(\Rightarrow y\le3\Rightarrow y_{max}=3\) khi \(cos2x=-1\)

Bài 1:

1: \(y=\frac{\sin x+2\cdot cosx+1}{2\cdot\sin x+cosx+3}\)

=>\(2y\cdot\sin x+y\cdot cosx+3y=\sin x+2\cdot cosx+1\)

=>\(\left(2y-1\right)\cdot\sin x+cosx\cdot\left(y-2\right)=1-3y\)

Để phương trình có nghiệm thì \(\left(2y-1\right)^2+\left(y-2\right)^2>=\left(1-3y\right)^2\)

=>\(4y^2-4y+1+y^2-4y+4\ge9y^2-6y+1\)

=>\(5y^2-8y+5-9y^2+6y-1\ge0\)

=>\(-4y^2-2y+4\ge0\)

=>\(y^2+\frac12y-1\le0\)

=>\(y^2+2\cdot y\cdot\frac14+\frac{1}{16}-\frac{17}{16}\le0\)

=>\(\left(y+\frac14\right)^2\le\frac{17}{16}\)

=>\(-\frac{\sqrt{17}}{4}\le y+\frac14\le\frac{\sqrt{17}}{4}\)

=>\(\frac{-\sqrt{17}-1}{4}\le y\le\frac{\sqrt{17}-1}{4}\)

=>\(y_{\min}=\frac{-\sqrt{17}-1}{4}\) và \(y_{\max}=\frac{\sqrt{17}-1}{4}\)

2: \(y=2\cdot\sin^2x-3\cdot\sin x\cdot cosx+cos^2x\)

\(=2\cdot\frac{1-cos2x}{2}-3\cdot\frac12\cdot\sin2x+\frac{1+cos2x}{2}\)

\(=1-cos2x-\frac32\cdot\sin2x+\frac12+\frac12\cdot cos2x\)

\(=-\frac32\cdot\sin2x-\frac12\cdot cos2x+\frac32=-\frac12\left(3\cdot\sin2x+cos2x-3\right)\)

\(=-\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\cdot\sin2x+\frac{1}{\sqrt{10}}\cdot cos2x-\frac{3}{\sqrt{10}}\right)\)

\(=-\frac{\sqrt{10}}{2}\cdot\left\lbrack\sin\left(2x+\alpha\right)-\frac{3}{\sqrt{10}}\right\rbrack\) , với \(cosa=\frac{3}{\sqrt{10}};\sin a=\frac{1}{\sqrt{10}}\)

\(=-\frac{\sqrt{10}}{2}\cdot\sin\left(2x+\alpha\right)+\frac32\)

Ta có: \(-1\le\sin\left(2x+a\right)\le1\)

=>\(-1\cdot\frac{-\sqrt{10}}{2}\ge\frac{-\sqrt{10}}{2}\sin\left(2x+a\right)\ge1\cdot\frac{-\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)\le\frac{\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}+\frac32\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)+\frac32\le\frac{\sqrt{10}}{2}+\frac32\)

=>\(y_{\min}=\frac{-\sqrt{10}+3}{2};y_{\max}=\frac{\sqrt{10}+3}{2}\)

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

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

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

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

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

y=(sin2x-3)^2-6

-1<=sin2x<=1

=>-4<=sin2x-3<=-2

=>4<=(sin2x-3)^2<=16

=>-2<=y<=10

y min khi sin2x-3=-2

=>sin 2x=1

=>2x=pi/2+k2pi

=>x=pi/4+kpi

y max khi sin 2x-3=-4

=>sin 2x=-1

=>2x=-pi/2+k2pi

=>x=-pi/4+kpi

\(y=4-\frac{5}{4}\left(2sin2x.cos2x\right)^2\)

\(y=4-\frac{5}{4}sin^24x\)

Do \(0\le sin^24x\le1\)

\(\Rightarrow\frac{11}{4}\le y\le4\)

\(y_{min}=\frac{11}{4}\) khi \(sin^24x=1\)

\(y_{max}=4\) khi \(sin^24x=0\)

cảm ơn ạ

c.

\(y=2sin2x-1\)

Do \(-1\le sin2x\le1\Rightarrow-3\le y\le1\)

\(y_{min}=-3\) khi \(sin2x=-1\)

\(y_{max}=1\) khi \(sin2x=1\)

d.

\(-1\le sin3x\le1\Rightarrow-1\le y\le3\)

e.

\(0\le sin^22x\le1\Rightarrow1\le y\le4\)

Em c.ơn ạ