Chọn A.

1/ ĐKXĐ: \(\cos2x\ne0\)

\(\frac{\cos4x}{\cos2x}=\frac{\sin2x}{\cos2x}\)\(\Leftrightarrow\cos4x-\sin2x=0\)

\(\Leftrightarrow2\cos^22x-1-\sin2x=0\)

\(\Leftrightarrow2-2\sin^22x-1-\sin2x=0\)

\(\Leftrightarrow2\sin^22x+\sin2x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\frac{1}{2}=\sin\frac{\pi}{6}\\\sin2x=-1=\sin\frac{-\pi}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{6}+2k\pi\\2x=\frac{5\pi}{6}+2k\pi\\2x=\frac{-\pi}{2}+2k\pi\left(l\right)\\2x=\frac{3\pi}{2}+2k\pi\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)

2/ \(\sin2.4x+\cos4x=1+2\sin2x.\cos\left(2x+4x\right)\)

\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+2\sin2x.\left(\cos2x.\cos4x-\sin2x.\sin4x\right)\)

\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+2\sin2x.\cos2x.\cos4x-2\sin^22x.\sin4x\)

\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+\sin4x.\cos4x-\sin4x+\cos4x.\sin4x\)

Đến đây bn tự giải nốt nhé, lm kiểu bthg thôi bởi vì đã quy về hết sin4x và cos4x r

1.

\(cos2x-3cosx+2=0\)

\(\Leftrightarrow2cos^2x-3cosx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

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

\(x=k2\pi\in\left[\dfrac{\pi}{4};\dfrac{7\pi}{4}\right]\Rightarrow\) không có nghiệm x thuộc đoạn

\(x=\pm\dfrac{\pi}{3}+k2\pi\in\left[\dfrac{\pi}{4};\dfrac{7\pi}{4}\right]\Rightarrow x_1=\dfrac{\pi}{3};x_2=\dfrac{5\pi}{3}\)

\(\Rightarrow P=x_1.x_2=\dfrac{5\pi^2}{9}\)

2.

\(pt\Leftrightarrow\left(cos3x-m+2\right)\left(2cos3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=\dfrac{1}{2}\left(1\right)\\cos3x=m-2\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=\pm\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\)

Ta có: \(x=\pm\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\in\left(-\dfrac{\pi}{6};\dfrac{\pi}{3}\right)\Rightarrow x=\pm\dfrac{\pi}{9}\)

Yêu cầu bài toán thỏa mãn khi \(\left(2\right)\) có nghiệm duy nhất thuộc \(\left(-\dfrac{\pi}{6};\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}m-2=0\\m-2=1\\m-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=2\\m=3\\m=1\end{matrix}\right.\)

TH1: \(m=2\)

\(\left(2\right)\Leftrightarrow cos3x=0\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\in\left(-\dfrac{\pi}{6};\dfrac{\pi}{3}\right)\Rightarrow x=\dfrac{\pi}{6}\left(tm\right)\)

\(\Rightarrow m=2\) thỏa mãn yêu cầu bài toán

TH2: \(m=3\)

\(\left(2\right)\Leftrightarrow cos3x=0\Leftrightarrow x=\dfrac{k2\pi}{3}\in\left(-\dfrac{\pi}{6};\dfrac{\pi}{3}\right)\Rightarrow x=0\left(tm\right)\)

\(\Rightarrow m=3\) thỏa mãn yêu cầu bài toán

TH3: \(m=1\)

\(\left(2\right)\Leftrightarrow cos3x=-1\Leftrightarrow x=\dfrac{\pi}{3}+\dfrac{k2\pi}{3}\in\left(-\dfrac{\pi}{6};\dfrac{\pi}{3}\right)\Rightarrow\left[{}\begin{matrix}x=\pm\dfrac{1}{3}\\x=-1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(\Rightarrow m=2\) không thỏa mãn yêu cầu bài toán

Vậy \(m=2;m=3\)

3.

\(2sin^2\dfrac{x}{4}-3cos\dfrac{x}{4}=0\)

\(\Leftrightarrow2cos^2\dfrac{x}{4}+3cos\dfrac{x}{4}-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\dfrac{x}{4}=\dfrac{1}{2}\\cos\dfrac{x}{4}=-2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\dfrac{4\pi}{3}+k8\pi\in\left[0;8\pi\right]\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4\pi}{3}\\x=\dfrac{20\pi}{3}\end{matrix}\right.\)

\(\Rightarrow T=\dfrac{4\pi}{3}+\dfrac{20\pi}{3}=8\pi\)

3.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)

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

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

câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ

1.

\(sin\left(sinx\right)=0\)

\(\Leftrightarrow sinx=k\pi\) (1)

Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)

\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)

Thế vào (1)

\(\Rightarrow sinx=0\Rightarrow x=n\pi\)

2.

\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)

Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)

\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\) 

\(\Rightarrow\) Không tồn tại k thỏa mãn

Pt vô nghiệm