Exercise:
Solve,$\sum\limits_{k=1}^{10}{\sin \left( 1+kx \right)=}0$ , $x\in \left( 0,2\pi \right)$
Solution: Let $C=\sum\limits_{k=1}^{10}{\sin \left( 1+kx \right)}$ multiply by $2\sin \frac{x}{2}$ both sides to get
$2\sin \frac{x}{2}C=\sum\limits_{k=1}^{10}{2\sin \left( 1+kx \right)\sin }\left( \frac{x}{2} \right)$
Notice that $-2\sin p\sin q=\cos \left( p+q \right)-\cos \left( p-q \right)\Leftrightarrow 2\sin p\sin q=\cos \left( p-q \right)-\cos \left( p+q \right)$
So $2\sin \left( 1+kx \right)\sin \left( \frac{x}{2} \right)=\cos \left( 1+kx-\frac{x}{2} \right)-\cos \left( 1+kx+\frac{x}{2} \right)$
$=\cos \left( 1+\frac{\left( 2k-1 \right)x}{2} \right)-\cos \left( 1+\frac{\left( 2k+1 \right)x}{2} \right)$
Hence $2\sin \frac{x}{2}C=\sum\limits_{k=1}^{10}{\cos \left( 1+\frac{\left( 2k-1 \right)x}{2} \right)-\sum\limits_{k=1}^{10}{\cos \left( 1+\frac{\left( 2k+1 \right)x}{2} \right)}}$
\[\Rightarrow 2\sin \frac{x}{2}C=\left\{ \cos \left( 1+\frac{x}{2} \right)+\cos \left( 1+\frac{3x}{2} \right)+\cos \left( 1+\frac{5x}{2} \right)+...+\cos \left( 1+\frac{19x}{2} \right) \right\}\]
$-\left\{ \cos \left( 1+\frac{3x}{2} \right)+\cos \left( 1+\frac{5x}{2} \right)+\cos \left( 1+\frac{7x}{2} \right)+....+\cos \left( 1+\frac{21x}{2} \right) \right\}$
$\Rightarrow 2\sin \frac{x}{2}C=\cos \left( 1+\frac{x}{2} \right)-\cos \left( 1+\frac{21x}{2} \right)$
Observe that, $p+q=1+\frac{x}{2}\,\,\,\And \,\,\,p-q=1+\frac{21x}{2}\Leftrightarrow p=1+\frac{11x}{2}\,\,\And \,\,q=-5x$
$\Rightarrow 2\sin \frac{x}{2}C=2\sin \left( 1+\frac{11x}{2} \right)\sin \left( -5x \right)$
Thus $C=\frac{\sin \left( -5x \right)}{\sin \frac{x}{2}}\sin \left( 1+\frac{11x}{2} \right)$
Therefore $\sum\limits_{k=1}^{10}{\sin \left( 1+kx \right)=\frac{-\sin 5x}{\sin \frac{x}{2}}\sin \left( 1+\frac{11x}{2} \right)=0}$
So $-\sin 5x\sin \left( 1+\frac{11x}{2} \right)=0\Leftrightarrow \sin 5x=0\,\,or\,\,\sin \left( 1+\frac{11x}{2} \right)=0$
$\Leftrightarrow \sin 5x=\sin \left( 0 \right)\,\,\,or\,\,\sin \left( 1+\frac{11x}{2} \right)=\sin 0$
So $5x=0+2k\pi \,\,\,or\,\,5x=\pi +2k\pi \,\,::\,\,\,1+\frac{11x}{2}=0+2k\pi \,\,or\,\,1+\frac{11x}{2}=\pi +2k\pi $
$x=\frac{2k\pi }{5}\,\,\,or\,\,x=\frac{\pi +2k\pi }{5}\,\,\,::\,\,\,\,\frac{11x}{2}=-1+2k\pi \,\,or\,\,\frac{11x}{2}=\pi -1+2k\pi $
$x=\left\{ \frac{\pi }{5},\frac{2\pi }{5},\frac{3\pi }{5},\frac{4\pi }{5},\pi ,\frac{6\pi }{5},\frac{7\pi }{5},\frac{8\pi }{5},\frac{9\pi }{5} \right\}\,\,=\frac{\left( k+1 \right)\pi }{5}\,\,,k=0,1,2,...,9$ so 9 roots
Or $x=\left\{ \frac{-2+2\pi }{11} \right.,\frac{-2+4\pi }{11},\frac{-2+6\pi }{11},\frac{-2+8\pi }{11},\frac{-2+10\pi }{11},\frac{-2+12\pi }{11},\frac{-2+14\pi }{11},\frac{-2+16\pi }{11},$
$\frac{-2+18\pi }{11},\frac{-2+20\pi }{11},\left. \frac{-2+22\pi }{11} \right\}=\frac{-2+2k\pi }{11}\,\,\,,\,\,k=1,2,...,11$ so 11 roots
Hence there are 20 roots in $\left( 0,2\pi \right)$ for this equation.
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