\(sinx\left(1+2cos2x+2cos4x+2cos6x\right)\)

\(=sinx+2sinx.cos2x+2sinx.cos4x+2sinx.cos6x\)

\(=sinx+sin3x+sin\left(-x\right)+sin5x+sin\left(-3x\right)+sin7x+sin\left(-5x\right)\)

\(=sinx+sin3x-sinx+sin5x-sin3x+sin7x-sin5x\)

\(=sin7x\)

\(A=\frac{cosx-cos3x+cos4x-cos2x}{sinx-sin3x+sin4x-sin2x}=\frac{2sin2x.sinx-2sin3x.sinx}{-2cos2x.sinx+2cos3x.sinx}\)

\(=\frac{sin2x-sin3x}{cos3x-cos2x}=\frac{-2cos\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}{-2sin\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}=cot\left(\frac{5x}{2}\right)\)

\(B=sinx+2cos2x.sinx+2cos4x.sinx+2cos6x.sinx\)

\(=sinx+sin3x-sinx+sin5x-sin3x+sin7x-sin5x\)

\(=sin7x\)

\(\Leftrightarrow2sin5x.sinx+1=2cos4x.sinx+2cos2x.sinx+3sinx\)

\(\Leftrightarrow2sin5x.sinx+1=sin5x-sin3x+sin3x-sinx+3sinx\)

\(\Leftrightarrow2sin5x.sinx-sin5x-2sinx+1=0\)

\(\Leftrightarrow sin5x\left(2sinx-1\right)-\left(2sinx-1\right)=0\)

1.

\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)

\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)

\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

2.

\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Nhận thấy \(sinx=0\) ko phải nghiệm

\(\Leftrightarrow sinx\left(2cos2x+1\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)

\(\Leftrightarrow\left(2cos2x.sinx+sinx\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)

\(\Leftrightarrow\left(sin3x-sinx+sinx\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)

\(\Leftrightarrow\left(2cos6x.sin3x+sin3x\right)\left(2cos18x+1\right)=sinx\)

\(\Leftrightarrow\left(2cos18x.sin9x+sin9x\right)=sinx\)

\(\Leftrightarrow sin27x=sinx\)

=>2cos2x=pi(loại) hoặc sin x-cosx=0

=>sin x-cosx=0

=>sin(x-pi/4)=0

=>x-pi/4=kpi

=>x=kpi+pi/4

mà x\(\in\left[-pi;pi\right]\)

nên \(x\in\left\{\dfrac{pi}{4};-\dfrac{3}{4}pi\right\}\)

=> D