Đặt \(t=tan\dfrac{x}{2}\Rightarrow\left\{{}\begin{matrix}t\in\left[0;1\right]\\sinx=\dfrac{2t}{1+t^2}\\cosx=\dfrac{1-t^2}{1+t^2}\end{matrix}\right.\)

Pt trở thành: \(\dfrac{m.2t}{1+t^2}+\dfrac{1-t^2}{1+t^2}=1\)

\(\Leftrightarrow2mt+1-t^2=1+t^2\)

\(\Leftrightarrow2mt-2t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=m\end{matrix}\right.\)

\(\Rightarrow\) Để pt có 2 nghiệm thuộc đoạn đã cho thì \(0< m\le1\)

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.

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

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

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

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

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

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

Đáp án A

Ta có sin 2 2 x + 3 sin 2 x + 2 = 0 ⇔ sin 2 x + 1 sin 2 x + 2 = 0 ⇔ sin 2 x = - 1 ⇔ 2 x = - π 2 + k 2 π ⇔ x = - π 4 + k π  mà x ∈ 0 ; 10 π suy ra 0 ≤ - π 4 + k π ≤ 10 π ⇔ 1 4 ≤ k ≤ 41 4 → k ∈ ℤ k = 1 ; 2 ; . . . ; 10  

Vậy tổng tất cả các nghiệm của phương trình là  T = 10 . - π 4 + 1 + 2 + . . . + 10 π = 105 2 π .

ĐKXĐ: \(cosx\ne-\dfrac{\sqrt{3}}{2}\) \(\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{5\pi}{6}+k2\pi\\x\ne\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(pt\Rightarrow3-\left(1-2sin^2x\right)+2sinx.cosx-5sinx-cosx=0\)

\(\Leftrightarrow2sin^2x-5sinx+2+cosx\left(2sinx-1\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx+cosx=2\left(vn\right)\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Loại nghiệm

\(\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)

\(0\le\dfrac{\pi}{6}+k2\pi\le2022\pi\Rightarrow0\le k\le1010\)

\(\Rightarrow\sum x=1011.\dfrac{\pi}{6}+2\pi\left(0+1+2+...+1010\right)=\dfrac{1011\pi}{6}+2\pi.\dfrac{1010.1011}{2}=...\)