\(\Leftrightarrow2cos4x.cosx+2cos^24x-1+1=0\)

\(\Leftrightarrow2cos4x\left(cos4x+cosx\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=cos\left(\pi-x\right)\end{matrix}\right.\)

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

A=\(\frac{\left(cos7x+cos10x\right)-\left(cos8x+cos9x\right)}{\left(sin7x+sin10x\right)-\left(sin8x+sin9x\right)}\) =\(\frac{2cos\frac{17x}{2}.cos\frac{3x}{2}-2cos\frac{17x}{2}.cos\frac{x}{2}}{2sin\frac{17x}{2}.cos\frac{3x}{2}-2sin\frac{17x}{2}.cos\frac{x}{2}}\)

=\(\frac{2cos\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}{2sin\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}\)=\(\frac{cos\frac{17x}{2}}{sin\frac{17x}{2}}\)=cotg\(\frac{17x}{2}\)

\(A=\frac{cos7x-cos8x-cos9x+cos10x}{sin7x-sin8x-sin9x+sin10x}=\frac{(cos10x+cos7x)-\left(cos9x+cos8x\right)}{\left(sin10x+sin7x\right)-\left(sin9x+sin8x\right)}.\) 

     \(=\frac{2cos\frac{17x}{2}cos\frac{3x}{2}-2cos\frac{17x}{2}cos\frac{x}{2}}{2sin\frac{17x}{2}cos\frac{3x}{2}-2sin\frac{17x}{2}cos\frac{x}{2}}=\frac{2cos\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}{2sin\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}=cotan\frac{17x}{2}.\)  

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

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

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

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

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

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HOK TỐT

\(\Leftrightarrow cos8x+cos2x+sinx=cos8x\)

\(\Leftrightarrow cos2x=-sinx\)

\(\Leftrightarrow cos2x=cos\left(x+\frac{\pi}{2}\right)\)

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

\(\Leftrightarrow...\)