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.

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

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

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

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

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

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

a/

\(cos^4x-\left(1-2sin^2x\right)+2sin^6x=0\)

\(\Leftrightarrow\left(cos^2x+1\right)\left(cos^2x-1\right)+2sin^2x\left(sin^4x+1\right)=0\)

\(\Leftrightarrow-sin^2x\left(cos^2x+1\right)+2sin^2x\left(sin^4x+1\right)=0\)

\(\Leftrightarrow sin^2x\left(2sin^4x-cos^2x+1\right)=0\)

\(\Leftrightarrow sin^2x\left(2sin^4x+sin^2x\right)=0\)

\(\Leftrightarrow sin^4x\left(2sin^2x+1\right)=0\)

\(\Leftrightarrow sinx=0\)

\(\Leftrightarrow x=k\pi\)

b/

\(cos4x=\frac{1}{2}+\frac{1}{2}cos6x\)

\(\Leftrightarrow2\left(2cos^22x-1\right)=1+4cos^32x-3cos2x\)

\(\Leftrightarrow4cos^32x-4cos^22x-3cos2x+3=0\)

\(\Leftrightarrow\left(cos2x-1\right)\left(4cos^22x-3\right)=0\)

\(\Leftrightarrow\left(cos2x-1\right)\left(2cos4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos4x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+\frac{k\pi}{2}\end{matrix}\right.\)

\(\Rightarrow x=\left\{0;-\frac{11\pi}{12};-\frac{5\pi}{12};\frac{\pi}{12};\frac{7\pi}{12};-\frac{7\pi}{12};-\frac{\pi}{12};\frac{5\pi}{12};\frac{11\pi}{12}\right\}\)

Bạn tự cộng lại

c/

\(\Leftrightarrow2cos^2x-1-\left(2m+1\right)cosx+m+1=0\)

\(\Leftrightarrow2cos^2x-\left(2m+1\right)cosx+m=0\)

\(\Leftrightarrow2cos^2x-cosx-2mcosx+m=0\)

\(\Leftrightarrow cosx\left(2cosx-1\right)-m\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left(cosx-m\right)\left(2cosx-1\right)=0\)

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

Do \(cosx=\frac{1}{2}\) vô nghiệm trên \(\left(\frac{\pi}{2};\frac{3\pi}{2}\right)\) nên pt có nghiệm khi và chỉ khi \(cosx=m\) có nghiệm trên khoảng đã cho

Mà \(-1< cosx< 0\Rightarrow-1< m< 0\)

\(\Leftrightarrow2\left(cos^2x-sin^2x\right)+sinx.cosx\left(sinx+cosx\right)=m\left(sinx+cosx\right)\)

\(\Leftrightarrow\left(2cosx-2sinx\right)\left(sinx+cosx\right)+sinx.cosx\left(sinx+cosx\right)=m\left(sinx+cosx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(\text{vô nghiệm trên đoạn xét}\right)\\2cosx-2sinx+sinx.cosx=m\left(1\right)\end{matrix}\right.\) 

Xét (1), đặt \(t=cosx-sinx=\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)\)

\(\Rightarrow\left\{{}\begin{matrix}t\in\left[-1;1\right]\\sinx.cosx=\dfrac{1-t^2}{2}\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2t+\dfrac{1-t^2}{2}=m\)

Xét hàm \(f\left(t\right)=-\dfrac{1}{2}t^2+2t+\dfrac{1}{2}\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=2\notin\left[-1;1\right]\) ; \(f\left(-1\right)=-2\) ; \(f\left(1\right)=2\)

\(\Rightarrow-2\le f\left(t\right)\le2\Rightarrow-2\le m\le2\)