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

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

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

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

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

a, \(y=2sin^2x-cos2x=1-2cos2x\)

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

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

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

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

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

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

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

\(y=sin^3x+2sin^2x+sinx-2\)

đặt \(t=sinx\) với \(t\in\left[-1;1\right]\)

 pt \(\Leftrightarrow\)\(y=t^3+2t^2+t-2\)

\(y'=3t^2+4t+1\)

\(y'=0\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=-\dfrac{1}{3}\end{matrix}\right.\)

vậy GTLN của y = 2 với t=1 \(\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

GTNN của y=-58/27  với \(t=-\dfrac{1}{3}\Leftrightarrow sinx=-\dfrac{1}{3}\Leftrightarrow x=sin^{-1}\left(-\dfrac{1}{3}\right)\)

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.

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

a.

Do \(-1\le sin\left(x+\frac{\pi}{6}\right)\le1\Rightarrow1\le y\le5\)

\(y_{min}=1\) khi \(sin\left(x+\frac{\pi}{6}\right)=1\)

\(y_{max}=5\) khi \(sin\left(x+\frac{\pi}{6}\right)=-1\)

b.

\(y=2\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]+3\)

\(y=2-4sin^2x.cos^2x+3=5-sin^22x\)

Do \(0\le sin^22x\le1\Rightarrow4\le y\le5\)

\(y_{min}=4\) khi \(sin^22x=1\)

\(y_{max}=5\) khi \(sin^22x=0\)

1: \(-1<=cosx\le1\)

=>\(-3\le-3\cdot cosx\le3\)

=>\(-3+5\le-3\cdot cosx+5\le3+5\)

=>2<=y<=8

y min=2 khi cosx=1

=>\(x=k2\pi\)

y min=8 khi cosx=-1

=>\(x=\pi+k2\pi\)

3: \(y=cos^2x+2\cdot cos2x\)

\(=\frac{1+cos2x}{2}+2\cdot cos2x=2,5\cdot cos2x+0,5\)

Ta có: \(-1\le cos2x\le1\)

=>\(-2,5\le2,5cos2x\le2,5\)

=>\(-2,5+0,5\le2,5cos2x+0,5\le2,5+0,5\)

=>-2<=y<=3

y min=-2 khi cos2x=-1

=>\(2x=\pi+k2\pi\)

=>\(x=\frac{\pi}{2}+k\pi\)

y max=3 khi cos2x=1

=>\(2x=k2\pi\)

=>\(x=k\pi\)

6: \(y=\sqrt3\cdot\sin x-cosx-2\)

\(=2\left(\frac{\sqrt3}{2}\cdot\sin x-\frac12\cdot cosx\right)-2=2\cdot\sin\left(x-\frac{\pi}{6}\right)-2\)

Ta có: \(-1\le\sin\left(x-\frac{\pi}{6}\right)\le1\)

=>\(-2\le2\sin\left(x-\frac{\pi}{6}\right)\le2\)

=>\(-2-2\le2\sin\left(x-\frac{\pi}{6}\right)-2\le2-2\)

=>-4<=y<=0

y min=-4 khi \(\sin\left(x-\frac{\pi}{6}\right)=-1\)

=>\(x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)

=>\(x=-\frac{\pi}{2}+\frac{\pi}{6}+k2\pi=-\frac26\pi+k2\pi=-\frac13\pi+k2\pi\)

y max=0 khi \(\sin\left(x-\frac{\pi}{6}\right)=1\)

=>\(x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac23\pi+k2\pi\)

Đáp án đúng : C

24.

\(cos\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)

\(y_{max}=4\)

26.

\(y=\sqrt{2}cos\left(2x-\dfrac{\pi}{4}\right)\)

Do \(cos\left(2x-\dfrac{\pi}{4}\right)\le1\Rightarrow y\le\sqrt{2}\)

\(y_{max}=\sqrt{2}\)

b.

\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)