ĐK: \(x\ne\dfrac{\pi}{4}+k\pi;x\ne\dfrac{k\pi}{2}\)

\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}=0\)

\(\Leftrightarrow2sin^2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sin^2x-1+cos4x-cos2x+1=0\)

\(\Leftrightarrow2cos^22x-2cos2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=k\pi\end{matrix}\right.\)

Đối chiếu điều kiện ta được \(x=-\dfrac{\pi}{4}+k\pi\)

- Hầu như các OLmers toàn tầm khoảng 2k4 đến 2k9 nên mk nghĩ là câu này của bn khó cs ai TL đc =))

- Mk nghĩ bn nên vào web : H để đăng bài ! Vì mk thấy ở đó chuyên giải mấy bài khó -,-

- Hoăc bn cs thể nhờ https://olm.vn/thanhvien/linhchi_nguyenthi1997 ( cj này là quản lý của olm và hay giải mấy bài khó )

Ckuc bn hok tốt =))

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

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

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

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

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

Đk:\(cosx\ne\dfrac{1}{2}\) \(\Rightarrow cosx\ne\pm\dfrac{\pi}{3}+k2\pi\);\(k\in Z\)

Pt \(\Leftrightarrow\dfrac{\left(2-\sqrt{3}\right)cosx-\left[1-cos\left(x-\dfrac{\pi}{2}\right)\right]}{2cosx-1}=1\)

\(\Rightarrow\left(2-\sqrt{3}\right)cosx-1+cos\left(\dfrac{\pi}{2}-x\right)=2cosx-1\)

\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)

\(\Leftrightarrow2sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\) (\(k\in Z\)) kết hợp với đk \(\Rightarrow x=\dfrac{2\pi}{3}+k2\pi\)(\(k\in Z\))

ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow x\ne\pm\dfrac{\pi}{3}+k2\pi\)

\(\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)-1=2cosx-1\)

\(\Leftrightarrow sinx-\sqrt{3}cosx=0\)

\(\Leftrightarrow tanx=\sqrt{3}\)

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

Kết hợp ĐKXĐ \(\Rightarrow x=-\dfrac{2\pi}{3}+k2\pi\)

f: \(cos7x-\sqrt3\cdot\sin7x-\sin x=\sqrt3\cdot cosx\)

=>\(\frac12\cdot cos7x-\frac{\sqrt3}{2}\cdot\sin7x=\frac12\cdot\sin x+\frac{\sqrt3}{2}\cdot cosx\)

=>\(\sin\left(\frac{\pi}{6}-7x\right)=\sin\left(x+\frac{\pi}{3}\right)\)

=>\(\left[\begin{array}{l}-7x+\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\ -7x+\frac{\pi}{6}=\pi-x-\frac{\pi}{3}+k2\pi=-x+\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}-7x-x=\frac{\pi}{3}-\frac{\pi}{6}+k2\pi=\frac{\pi}{6}+k2\pi\\ -7x+x=\frac23\pi-\frac{\pi}{6}+k2\pi=\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}-8x=\frac{\pi}{6}+k2\pi\\ -6x=\frac12\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{\pi}{48}-\frac{k\pi}{4}\\ x=-\frac{1}{12}\pi-\frac{k\pi}{3}\end{array}\right.\)

e: \(5\cdot\sin2x-6\cdot cos^2x=13\)

=>\(5\cdot\sin2x-6\cdot\frac{1+cos2x}{2}=13\)

=>\(5\cdot\sin2x-3-3\cdot cos2x=13\)

=>\(5\cdot\sin2x-3\cdot cos2x=16\)

Vì \(5^2+\left(-3\right)^2=25+9=34<16^2\)

nên phương trình vô nghiệm

TH1 cosx=0 \(\Leftrightarrow1=0\left(vl\right)\)

TH2 \(cosx\ne0\) chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow-2tan^3x-6+1+tân^2x+3tanx\left(1+tan^2x\right)=0\)

\(\Leftrightarrow tan^3x+tan^2x+3tanx-5=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x+2tanx+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tan^2x+2tanx+5=0\left(VN\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)

Đặt \(\dfrac{x}{4}=t\)

\(2sin^22t-3cost=0\)

\(\Leftrightarrow8sin^2t.cos^2t-3cost=0\)

\(\Leftrightarrow8cos^2t\left(1-cos^2t\right)-3cost=0\)

\(\Leftrightarrow-8cos^4t+8cos^2t-3cost=0\)

\(\Leftrightarrow-cost\left(8cos^3t-8cost+3\right)=0\)

\(\Leftrightarrow cost\left(2cost-1\right)\left(4cos^2t+2cost-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\cost=\dfrac{1}{2}\\cost=\dfrac{-1+\sqrt{13}}{4}\\cost=\dfrac{-1-\sqrt{13}}{4}< -1\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\Leftrightarrow2cos2x.cos\left(\dfrac{\pi}{6}\right)-2sin2x.sin\left(\dfrac{\pi}{6}\right)+2sin2x-1=0\)

\(\Leftrightarrow\sqrt{3}cos2x+sin2x=1\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x=\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)