Đáp án là A

Chọn A

\(\Leftrightarrow cos6x-cos8x+2\left(1-cos4x\right)^2+\sqrt{3}sin6x=4-4cos4x\)

\(\Leftrightarrow cos6x-cos8x+2\left(1+cos^24x-2cos4x\right)+\sqrt{3}sin6x=4-4cos4x\)

\(\Leftrightarrow cos6x-cos8x+cos8x+3-4cos4x+\sqrt{3}sin6x=4-4cos4x\)

\(\Leftrightarrow cos6x+\sqrt{3}sin6x=1\)

\(\Leftrightarrow cos\left(6x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow...\)

\(\Leftrightarrow1-cos4x+sin7x-1=sinx\)

\(\Leftrightarrow sin7x-sinx-cos4x=0\)

\(\Leftrightarrow2.cos4x.sin3x-cos4x=0\)

\(\Leftrightarrow cos4x\left(2.sin3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\3x=\dfrac{\pi}{6}+k2\pi\\3x=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\)) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\) (\(k\in Z\))

Kết luận:...

\(\Leftrightarrow2cos5x.cosx=2cos5x.sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos5x=0\\cosx=sin2x=cos\left(\frac{\pi}{2}-2x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{2}-2x+k2\pi\\x=2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{10}+\frac{k\pi}{5}\\x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

a, cos2x - sin7x = 0

⇔ cos2x = sin7x

⇔ cos2x = cos \(\left(7x-\dfrac{\pi}{2}\right)\)

⇔ \(\left[{}\begin{matrix}7x-\dfrac{\pi}{2}=2x+k2\pi\\7x-\dfrac{\pi}{2}=-2x+k2\pi\end{matrix}\right.\) với k là số nguyên

⇔ \(\left[{}\begin{matrix}x=\dfrac{\pi}{10}+\dfrac{k.2\pi}{5}\\x=\dfrac{\pi}{18}+\dfrac{k2\pi}{9}\end{matrix}\right.\) với k là số nguyên 

a/

\(\Leftrightarrow2sin4x.cos3x=2sin7x.cos3x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\sin7x=sin4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\frac{\pi}{2}+k\pi\\7x=4x+k2\pi\\7x=\pi-4x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k\pi}{3}\\x=\frac{k2\pi}{3}\\x=\frac{\pi}{11}+\frac{k2\pi}{11}\end{matrix}\right.\)

b.

\(\Leftrightarrow2cos4x.cosx=2cos8x.cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos8x=cos4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=4x+k2\pi\\8x=-4x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{k\pi}{2}\\x=\frac{k\pi}{6}\end{matrix}\right.\) \(\Leftrightarrow x=\frac{k\pi}{6}\)