a/ \(y=2cos\left(\frac{\pi}{14}\right)cos\left(x-\frac{\pi}{14}\right)\)

Do \(-1\le cos\left(x-\frac{\pi}{14}\right)\le1\) với mọi x

\(\Rightarrow-2cos\left(\frac{\pi}{14}\right)\le y\le2cos\left(\frac{\pi}{14}\right)\)

\(y_{min}=-2cos\left(\frac{\pi}{14}\right)\) khi \(cos\left(x-\frac{\pi}{14}\right)=-1\)

\(y_{max}=2cos\left(\frac{\pi}{14}\right)\) khi \(cos\left(x-\frac{\pi}{14}\right)=1\)

b/ \(y=\sqrt{3}cos2x-\frac{1}{2}sin2x=\frac{\sqrt{13}}{2}\left(\frac{2\sqrt{39}}{13}cos2x-\frac{\sqrt{13}}{13}sin2x\right)\)

\(\Rightarrow y=\frac{\sqrt{13}}{2}cos\left(2x+a\right)\) với \(a\in\left(0;\pi\right)\) sao cho \(cosa=\frac{2\sqrt{39}}{13}\)

Do \(-1\le cos\left(2x+a\right)\le1\Rightarrow-\frac{\sqrt{13}}{2}\le y\le\frac{\sqrt{13}}{2}\)

c/ \(y=4sin^2x+4sinx+1+4cos^2x-4\sqrt{3}cosx+3\)

\(=8+4sinx-4\sqrt{3}cosx=8+8\left(\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx\right)\)

\(=8+8sin\left(x-\frac{\pi}{3}\right)\)

Do \(-1\le sin\left(x-\frac{\pi}{3}\right)\le1\Rightarrow0\le y\le16\)

\(y=1-sin^2x+sinx+1=-sin^2x+sinx+2\)

\(\Rightarrow y=-\left(sinx-\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)

\(\Rightarrow y_{max}=\frac{9}{4}\) khi \(sinx=\frac{1}{2}\)

\(y=\left(sinx+1\right)\left(2-sinx\right)\)

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

\(\Rightarrow y=\left(sinx+1\right)\left(2-sinx\right)\ge0\)

\(\Rightarrow y_{min}=0\) khi \(sinx=-1\)

đọc lại lý thuyết rồi làm 

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 \(\sin\left(x-\frac{\pi}{6}\right)=1\)

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

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

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

\(y_{min}=-1\) khi \(sin3x=-1\)

\(y_{max}=3\) khi \(sin3x=1\)

b/ \(0\le cos^22x\le1\Rightarrow1\le y\le2\)

\(y_{min}=1\) khi \(cos^22x=0\)

\(y_{max}=3\) khi \(cos^22x=1\)

c/ \(y=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)+2\Rightarrow-\sqrt{2}+2\le y\le\sqrt{2}+2\)

\(y_{min}=-\sqrt{2}+2\) khi \(sin\left(x+\frac{\pi}{4}\right)=-1\)

\(y_{max}=\sqrt{2}+2\) khi \(sin\left(x+\frac{\pi}{4}\right)=1\)

d/ \(y=3cosx-\left(2cos^2x-1\right)+5=-2cos^2x+3cosx+6\)

\(y=-2\left(cosx-\frac{3}{4}\right)^2+\frac{57}{8}\le\frac{57}{8}\)

\(y_{max}=\frac{57}{8}\) khi \(cosx=\frac{3}{4}\)

\(y=\left(cosx+1\right)\left(-2cosx+5\right)+1\ge1\)

\(y_{min}=1\) khi \(cosx=-1\)