A =

= tan 3x

\(A=\dfrac{sinx+sin3x+sin2x}{cosx+cos3x+cos2x}=\dfrac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\dfrac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=tan2x\)

a2: \(2\cdot\sin17x+\sqrt3\cdot cos5x+\sin5x=0\)

=>\(\sin17x+\frac{\sqrt3}{2}\cdot cos5x+\frac12\cdot\sin5x=0\)

=>\(\sin17x+\sin\left(5x+\frac{\pi}{3}\right)=0\)

=>\(\sin17x=-\sin\left(5x+\frac{\pi}{3}\right)=\sin\left(-5x-\frac{\pi}{3}\right)\)

=>\(\left[\begin{array}{l}17x=-5x-\frac{\pi}{3}+k2\pi\\ 17x=\pi+5x+\frac{\pi}{3}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}22x=-\frac{\pi}{3}+k2\pi\\ 12x=\frac43\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{1}{66}\pi+\frac{k\pi}{11}\\ x=\frac19\pi+\frac{k\pi}{6}\end{array}\right.\)

a3: \(cos7x-\sin5x=\sqrt3\left(cos5x-\sin7x\right)\)

=>\(cos7x+\sqrt3\cdot\sin7x=\sqrt3\cdot cos5x+\sin5x\)

=>\(\frac{\sqrt3}{2}\cdot\sin7x+\frac12\cdot cos7x=\frac{\sqrt3}{2}\cdot cos5x+\frac12\cdot\sin5x\)

=>\(\sin\left(7x+\frac{\pi}{6}\right)=\sin\left(5x+\frac{\pi}{3}\right)\)

=>\(\left[\begin{array}{l}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\ 7x+\frac{\pi}{6}=\pi-5x-\frac{\pi}{3}+k2\pi=-5x+\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}-\frac{\pi}{6}+k2\pi=\frac{\pi}{6}+k2\pi\\ 12x=\frac23\pi-\frac{\pi}{6}+k2\pi=\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{12}+k\pi\\ x=\frac{1}{24}\pi+\frac{k\pi}{6}\end{array}\right.\)

Chọn C.

Ta có:

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

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

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

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