\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)

\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)

\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)

\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)

`cos 3x+cos 7x=sin 3x-sin 7x`

`<=>sin 3x-cos 3x=sin 7x+cos 7x`

`<=>sin(3x-\pi/4)=sin(7x+\pi/4)`

`<=>[(7x+\pi/4=3x-\pi/4+k2\pi),(7x+\pi/4=[3\pi]/4-3x+k2\pi):}`

`<=>[(x=-\pi/8+[k\pi]/2),(x=\pi/20+[k\pi]/5):}`

\(A=cosx+cos3x+cos5x+cos7x\)

\(=2cos4x.cos3x+2cos4x.cosx\)

\(=2cos4x.\left(cos3x+cosx\right)\)

\(=4cos4x.cos2x.cosx\)

\(D=\frac{1+sin2x+cos2x}{1+sin2x-cos2x}=\frac{1+2sinxcosx+2cos^2x-1}{1+2sinxcosx-1+2sin^2x}\)

\(D=\frac{cosx\left(sinx+cosx\right)}{sinx\left(sinx+cosx\right)}=cotx\)

\(F=\frac{sinx+sin4x+sin7x}{cosx+cos4x+cos7x}\)

\(F=\frac{2sin4xcos3x+sin4x}{2cos4xcos3x+cos4x}\)

\(F=\frac{2sin4x\left(cos3x+1\right)}{2cos4x\left(cos3x+1\right)}=tan4x\)

\(G=\frac{cos2x-sin4x-cos6x}{cos2x+sin4x-cos6x}=\frac{-2sin4xsin2x-sin4x}{-2sin4xsin2x+sin4x}\)

\(G=\frac{-sin4x\left(2sin2x+1\right)}{-sin4x\left(2sin2x-1\right)}=\frac{2sin2x+1}{2sin2x-1}\)

\(sin\dfrac{3x}{2}\left(cosx+cos4x+cos7x\right)\)

\(=\)\(sin\dfrac{3x}{2}.cosx+sin\dfrac{3x}{2}.cos4x+sin\dfrac{3x}{2}.cos7x=\dfrac{1}{2}\left[sin\dfrac{x}{2}+sin\dfrac{5x}{2}\right]+\dfrac{1}{2}\left[sin\left(-\dfrac{5x}{2}\right)+sin\dfrac{11x}{2}\right]+\dfrac{1}{2}\left[sin\left(-\dfrac{11x}{2}\right)+sin\dfrac{17x}{2}\right]\)

\(=\dfrac{1}{2}\left(sin\dfrac{x}{2}+sin\dfrac{17x}{2}\right)\)\(=\dfrac{1}{2}.2.sin\dfrac{9x}{2}.cos4x=sin\dfrac{9x}{2}.cos4x\) 

\(sin\dfrac{3x}{2}\left(sinx+sin4x+sin7x\right)\)

\(=sin\dfrac{3x}{2}.sinx+sin\dfrac{3x}{2}.sin4x+sin\dfrac{3x}{2}.sin7x\)\(=\dfrac{1}{2}\left(cos\dfrac{x}{2}-cos\dfrac{5x}{2}\right)+\dfrac{1}{2}\left(cos\dfrac{-5x}{2}-cos\dfrac{11x}{2}\right)+\dfrac{1}{2}\left(cos\dfrac{-11x}{2}-cos\dfrac{17x}{2}\right)\)

\(=\dfrac{1}{2}\left(cos\dfrac{x}{2}-cos\dfrac{17x}{2}\right)\)\(=\dfrac{1}{2}.-2.sin\dfrac{9x}{2}.sin\left(-4x\right)=sin\dfrac{9x}{2}.sin4x\)

Có \(\dfrac{cos7x+cos4x+cosx}{sin7x+sin4x+sinx}\)

\(=\dfrac{sin\dfrac{3x}{2}\left(cos7x+cos4x+cosx\right)}{sin\dfrac{3x}{2}\left(sin7x+sin4x+sinx\right)}\)\(=\dfrac{sin\dfrac{9x}{2}.cos4x}{sin\dfrac{9x}{2}.sin4x}\)\(=\dfrac{cos4x}{sin4x}\)

\(\Rightarrow\dfrac{cos4x}{sin4x}=\dfrac{1}{2}\)

\(\Leftrightarrow2cos4x=sin4x\)

\(\Leftrightarrow4.cos^24x=sin^24x\)

\(\Leftrightarrow4.cos^24x=1-cos^24x\)\(\Leftrightarrow cos^24x=\dfrac{1}{5}\Leftrightarrow\dfrac{1+cos8x}{2}=\dfrac{1}{5}\)

\(\Leftrightarrow cos8x=-\dfrac{3}{5}\)

Vậy..

a: ĐKXĐ: sin 2x<>1

=>2x<>pi/2+k2pi

=>x<>pi/4+kpi

\(\dfrac{cos2x}{sin2x-1}=0\)

=>cos2x=0

=>2x=pi/2+kpi

=>x=pi/4+kpi/2

Kết hợp ĐKXĐ, ta được:

x=3/4pi+k2pi hoặc x=7/4pi+k2pi

b: cos(sinx)=1

=>sin x=kpi

=>sin x=0

=>x=kpi

c: \(2\cdot sin^2x-1+cos3x=0\)

=>cos3x+cos2x=0

=>cos3x=-cos2x=-sin(pi/2-2x)=sin(2x-pi/2)

=>cos3x=cos(pi/2-2x+pi/2)=cos(pi-2x)

=>3x=pi-2x+k2pi hoặc 3x=-pi+2x+k2pi

=>x=-pi+k2pi hoặc x=pi/5+k2pi/5

e: cos3x=-cos7x

=>cos3x=cos(pi-7x)

=>3x=pi-7x+k2pi hoặc 3x=-pi+7x+k2pi

=>x=pi/10+kpi/5 hoặc x=pi/4-kpi/2

a. cos2x + cos4x + cos6x = 0

\(\Leftrightarrow\left(cos2x+cos6x\right)+cos4x=0\\ \Leftrightarrow2cos4x.cos2x+cos4x=0\\ \Leftrightarrow cos4x\left(2cos2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\left(k\in Z\right)}\)

1.

\(cos2x+cos6x+cos4x=0\)

\(\Leftrightarrow2cos4x.cos2x+cos4x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\2x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

2.

\(\Leftrightarrow1+cos2x+cosx+cos3x=0\)

\(\Leftrightarrow1+2cos^2x-1+2cos2x.cosx=0\)

\(\Leftrightarrow cos^2x+cos2x.cosx=0\)

\(\Leftrightarrow cosx\left(cos2x+cosx\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=-1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\frac{\left(sin3x+cosx\right)sin3x+\left(cos3x+sinx\right)cos3x}{cos4x}\)

\(=\frac{sin^23x+sin3x.cosx+cos^23x+cos3x.sinx}{cos4x}=\frac{1+sin3x.cosx+cos3x.sinx}{cos4x}\)

\(=\frac{1+sin4x}{cos4x}=\frac{sin^22x+cos^22x+2sin2x.cos2x}{cos^22x-sin^22x}=\frac{\left(cos2x+sin2x\right)^2}{\left(cos2x-sin2x\right)\left(cos2x+sin2x\right)}\)

\(=\frac{cos2x+sin2x}{cos2x-sin2x}=\frac{1+\frac{sin2x}{cos2x}}{1-\frac{sin2x}{cos2x}}=\frac{1+tan2x}{1-tan2x}\)