Chọn D

Ta sẽ biến đổi phương trình thành dạng tích

Chú ý: có thể dùng 4 đáp án thay vào phương trình để kiểm tra đâu là nghiệm

`A=[sin x+sin 2x+sin 3x]/[cos x+cos 2x+cos 3x]`

`A=[(sin x+sin 3x)+sin 2x]/[(cos x+cos 3x)+cos 2x]`

`A=[2sin 2x.cos (-x)+sin 2x]/[2cos 2x.cos (-x)+cos 2x]`

`A=[sin 2x(2cos(-x)+1)]/[cos 2x(2cos(-x)+1)]`

`A=[sin 2x]/[cos 2x]=tan 2x`.

`A=[sin x + sin 2x + sin 3x]/[cos x + cos 2x + cos 3x]`

`A=[2sin2x.cosx+sin2x]/[2cos2x.cosx+cos2x]`

`A=[sin2x(2cosx+1)]/[cos2x(2cosx+1)]`

`A=tan 2x`

\(A=\dfrac{sinx-sin2x+sin3x}{cosx-cos2x+cos3x}\)

\(ĐK\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne\dfrac{1}{2}\end{matrix}\right.\)  \(\Leftrightarrow\)  \(A=\dfrac{sinx+sin3x-sin2x}{cosx+cos3x-cos2x}\)     

\(\Leftrightarrow\)  \(\left\{{}\begin{matrix}=\dfrac{2sin2x.cosx-sin2x}{2cos2x.cosx-cos2x}\\=\dfrac{sin2x\left(2cosx-1\right)}{cos2x\left(2cosx-1\right)}\end{matrix}\right.\)  \(\Rightarrow\) \(A=tan2x\)

ĐKXĐ: ..

\(\frac{sin3x+sinx+sin2x}{cos3x+cosx+cos2x}=\sqrt{3}\)

\(\Leftrightarrow\frac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\sqrt{3}\)

\(\Leftrightarrow\frac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=\sqrt{3}\)

\(\Leftrightarrow tan2x=\sqrt{3}\)

\(\Leftrightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)

\(\Leftrightarrow sin2x\cdot sinx-cos2x\cdot sinx+sin2x\cdot cosx+sinx\cdot cos2x=cosx\left(sinx+cosx\right)\)

=>\(sin2x\left(sinx+cosx\right)=cosx\left(sinx+cosx\right)\)

=>\(\left(sinx+cosx\right)\cdot\left(sin2x-cosx\right)=0\)

=>\(cosx\cdot\left(2sinx-1\right)\cdot\sqrt{2}\cdot sin\left(x+\dfrac{pi}{4}\right)=0\)

=>\(\left[{}\begin{matrix}cosx=0\\2sinx-1=0\\sin\left(x+\dfrac{pi}{4}\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\sinx=\dfrac{1}{2}\\x+\dfrac{pi}{4}=kpi\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\x=-\dfrac{pi}{4}+kpi\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\x=-\dfrac{pi}{4}+kpi\\x=\dfrac{pi}{6}+k2pi\\x=\dfrac{5}{6}pi+k2pi\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{pi}{2}+kpi;-\dfrac{pi}{4}+kpi;\dfrac{pi}{6}+k2pi;\dfrac{5}{6}pi+k2pi\right\}\)

\(A=\frac{sin3x-sinx+cos2x}{cosx-cos3x+sin2x}=\frac{2cos2x.sinx+cos2x}{2sin2x.sinx+sin2x}=\frac{cos2x\left(2sinx+1\right)}{sin2x\left(2sinx+1\right)}=\frac{cos2x}{sin2x}=cot2x\)