\(2\sin2x-\cos2x=7\sin x+2\cos x-4\)

\(\Rightarrow4\sin x\cos x-\left(1-2\sin^2x\right)-7\sin x-2\cos x+4=0\)

\(\Rightarrow2\cos x\left(2\sin x-1\right)+\left(2\sin^2-7\sin x+3\right)=0\)

\(\Rightarrow2\cos x\left(2\sin x-1\right)+\left(2\sin x-1\right)\left(\sin x-3\right)=0\)

\(\Rightarrow\left(2\sin x-1\right)\left(2\cos x+\sin x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2\sin x-1=0\\2\cos x+\sin x=3,\left(vn\right)\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{\eta}{6}+k2\eta\\x=\frac{5\eta}{6}+k2\eta\end{cases}}}\)

\(\eta=Pi;3.14159\)

\(M=\dfrac{3sinx-2cosx}{5cosx+7sinx}\)

\(tanx=\dfrac{sinx}{cosx}\)

\(\Rightarrow M=\dfrac{3sinx}{\dfrac{cosx}{\dfrac{5cosx}{cosx}}}-\dfrac{2cosx}{\dfrac{cosx}{\dfrac{7sinx}{cosx}}}\)

\(M=\dfrac{3tanx-2}{5+7tanx}\)

\(M=\dfrac{3.2-2}{5+7.2}\)

\(M=\dfrac{4}{19}\)

a) <=> 4sinxcosx -(2cos²x-1)=7sinx+2cosx-4

<=> 2cos²x+(2-4sinx)cosx+7sinx-5=0

- sinx=1 => 2cos²x-2cosx+2=0 

pt trên vn

b) <=> 2sinxcosx-1+2sin²x+3sinx-cosx-1=0

<=> cos(2sinx-1)+2sin²x+3sinx-2=0

<=> cosx(2sinx-1)+(2sinx-1)(sinx+2)=0

<=> (2sinx-1)(cosx+sinx+2)=0

<=> sinx=1/2 hoặc cosx+sinx=-2(vn)

<=> x= \(\frac{\pi}{6}+k2\pi\) hoặc \(x=\frac{5\pi}{6}+k2\pi\left(k\in Z\right)\)

a/ ĐKXĐ: \(cosx\ne-\frac{1}{2}\)

\(\Leftrightarrow2cosx-1=6cosx+3\)

\(\Leftrightarrow4cosx=-4\Rightarrow cosx=-1\)

\(\Rightarrow x=\pi+k2\pi\)

b/

\(\Leftrightarrow cosx\left(2cos2x-1\right)-3cosx=0\)

\(\Leftrightarrow cosx\left(2cos2x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=2\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=\frac{\pi}{2}+k\pi\)

c/

\(\Leftrightarrow2sin2x.cos2x=0\)

\(\Leftrightarrow sin4x=0\)

\(\Rightarrow4x=k\pi\Rightarrow x=\frac{k\pi}{4}\)

tanx=sinx/cosx=5/3

=sin=3/5cos

=>2cosx-3(3/5cos)/3cosx+4(5/3cos)

=1/30

a, ĐK: \(x\ne\dfrac{5\pi}{6}+k2\pi;x\ne\dfrac{\pi}{6}+k2\pi\)

\(\dfrac{2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)}{2sinx-1}=-1\)

\(\Leftrightarrow2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)=1-2sinx\)

\(\Leftrightarrow-cos\left(3x-\dfrac{\pi}{2}\right)+\sqrt{3}cos^3x.\dfrac{cos^2x-3sin^2x}{cos^2x}=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(cos^2x-3sin^2x\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(4cos^2x-3\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cos3x=-2sinx\)

\(\Leftrightarrow\dfrac{1}{2}sin3x-\dfrac{\sqrt{3}}{2}cos3x-sinx=0\)

\(\Leftrightarrow sin\left(3x-\dfrac{\pi}{3}\right)-sinx=0\)

\(\Leftrightarrow2cos\left(2x-\dfrac{\pi}{6}\right)sin\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)

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

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

Đối chiếu điều kiện ta được:

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=\dfrac{7\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

Chọn đáp án D.