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

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

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

Đề là \(y=2sin^2x+cos^22x\) hả bạn? Và tìm GTNN, GTLN hay tìm tập giá trị?

e/

\(\Leftrightarrow3\left(1-cos6x\right)-\left(2cos^26x-1\right)=4\)

\(\Leftrightarrow-2cos^26x-3cos6x=0\)

\(\Leftrightarrow cos6x\left(2cos6x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cos6x=0\\cos6x=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow6x=\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=\frac{\pi}{12}+\frac{k\pi}{3}\)

d/

\(\Leftrightarrow3\left(1-cos2x\right)-2\left(1-cos^22x\right)=5\)

\(\Leftrightarrow2cos^22x-3cos2x-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\frac{3+\sqrt{41}}{4}\left(l\right)\\cos2x=\frac{3-\sqrt{41}}{4}\end{matrix}\right.\)

\(\Rightarrow x=\pm\frac{1}{2}arccos\left(\frac{3-\sqrt{41}}{4}\right)+k\pi\)

Nghiệm xấu quá :(

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

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

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

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

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