`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`

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

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

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

`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 =

= tan 3x

A =

= tan 3x

a: \(\sin3x+cos2x=1+2\cdot\sin x\cdot cos2x\)

=>sin3x+cos2x=1+sin(x+2x)+sin(x-2x)

=>sin3x+cos2x=1+sin3x-sin x

=>cos2x-1+sin x=0

=>\(1-2\cdot\sin^2x-1+\sin x=0\)

=>\(-2\cdot\sin^2x+\sin x=0\)

=>sin x(2sin x-1)=0

TH1: sin x=0

=>\(x=k\pi\)

TH2: 2sin x-1=0

=>\(\sin x=\frac12\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\)

b: \(\sin^3x+cos^3x=2\cdot\left(\sin^5x+cos^5x\right)\)

=>\(\sin^3x-2\cdot\sin^5x+cos^3x-2\cdot cos^5x=0\)

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

=>\(\sin^3x\cdot cos2x-cos^3x\cdot cos2x=0\)

=>\(cos2x\left(\sin^3x-cos^3x\right)=0\)

TH1: cos2x=0

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

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

TH2: \(\sin^3x-cos^3x=0\)

=>\(\sin^3x=cos^3x\)

=>sin x=cosx

=>\(\sin x-cosx=0\)

=>\(\sqrt2\cdot\sin\left(x-\frac{\pi}{4}\right)=0\)

=>\(\sin\left(x-\frac{\pi}{4}\right)=0\)

=>\(x-\frac{\pi}{4}=k\pi\)

=>\(x=\frac{\pi}{4}+k\pi\)

f: ĐKXĐ: \(\begin{cases}\sin x<>0\\ cosx<>0\end{cases}\Rightarrow\begin{cases}x<>k\pi\\ x<>\frac{\pi}{2}+k\pi\end{cases}\Rightarrow x<>\frac{k\pi}{2}\)

\(\frac{\tan x-\sin x}{\sin^3x}=\frac{1}{cosx}\)

=>\(\frac{\frac{\sin x}{cosx}-\sin x}{\sin^3x}=\frac{1}{cosx}\)

=>\(\frac{\frac{1}{cosx}-1}{\sin^2x}=\frac{1}{cosx}\)

=>\(\sin^2x=cosx\cdot\left(\frac{1}{cosx}-1\right)=1-cosx\)

=>\(1-cos^2x=1-cosx\)

=>\(cos^2x-cosx=0\)

=>cosx(cosx-1)=0

TH1: cosx=0

=>\(x=\frac{\pi}{2}+k\pi\) (loại)

TH2: cosx-1=0

=>cosx=1

=>\(x=k2\pi\)

=>sin x=0

=>Loại

\(cos^2x-\left(2sin\frac{x}{2}cos\frac{x}{2}\right)^2=cos^2x-sin^2x=cos2x\)

\(\frac{sin3x}{sinx}-\frac{cos3x}{cosx}=\frac{sin3x.cosx-cos3x.sinx}{sinx.cosx}=\frac{sin\left(3x-x\right)}{\frac{1}{2}sin2x}=\frac{2sin2x}{sin2x}=2\)

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

\(=\frac{cos2x+cos3x}{sin2x+sin3x}=\frac{2cos\frac{x}{2}.cos\frac{5x}{2}}{2sin\frac{5x}{2}.cos\frac{x}{2}}=cot\frac{5x}{2}\)