`\sin(7x)-\sin(2x)=\cos(2x)-\cos(7x)`
` <=> 2\cos(\frac{7x+2x}{2})\sin(\frac{7x-2x}{2})=-2\sin(\frac{2x+7x}{2})\sin(\frac{2x-7x}{2})`
`<=> 2\cos(\frac{9x}{2})\sin(\frac{5x}{2})=-2\sin(\frac{9x}{2})\sin(-\frac{5x}{2})`
`<=> 2\cos(\frac{9x}{2})\sin(\frac{5x}{2})=2\sin(\frac{9x}{2})\sin(\frac{5x}{2})`
`<=> \cos(\frac{9x}{2})\sin(\frac{5x}{2})=\sin(\frac{9x}{2})\sin(\frac{5x}{2})`
`<=> \cos(frac{9x}{2})\sin((5x)/2)-\sin((9x)/2)\sin((5x)/2)=0`
`<=>\sin((5x)/2)[\cos((9x)/2)-\sin((9x)/2)]`
`<=>` \(\left[ \begin{array}{l}\sin\dfrac{5x}{2}=0\\\cos\dfrac{9x}{2}-\sin\dfrac{9x}{2}=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}\dfrac{5x}{2}=k\pi,\quad k\in \mathbb{Z}\\-\sin\dfrac{9x}{2}=-\cos\dfrac{9x}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}\dfrac{5x}{2}=k\pi,\quad k\in \mathbb{Z}\\-\tan(\dfrac{9}{2}x)=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{2k\pi}{5}\\\dfrac{9}{2}x=\arctan(1)\end{array} \right. k\in\mathbb{Z}\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{2k\pi}{5}\\\dfrac{9}{2}x=\dfrac{\pi}{4}\end{array} \right.k\in\mathbb{Z}\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{2k\pi}{5}\\\dfrac{9}{2}x=\dfrac{\pi}{4}+k\pi\end{array} \right.k\in\mathbb{Z}\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{2k\pi}{5}\\18x =\pi +4k\pi\end{array} \right.k\in\mathbb{Z}\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{2k\pi}{5}\\x=\dfrac{\pi+4k\pi}{18}\end{array} \right.\quad k\in\mathbb{Z}\)
