The number of solution of the equation,sumlim | vedclass

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Question

$cos \,x + cos\, 2x + cos \,3x + cos \,4x + cos\, 5 x = 0$

$2\, cos \,3x \,cos \,2x + 2\, cos \,3x \,cos\, x + cos \,3x = 0$

$cos\, 3x [2 \,cos

Vedclass · Accepted Answer

$5$

Vedclass · Answer

$cos \,x + cos\, 2x + cos \,3x + cos \,4x + cos\, 5 x = 0$

$2\, cos \,3x \,cos \,2x + 2\, cos \,3x \,cos\, x + cos \,3x = 0$

$cos\, 3x [2 \,cos \,2x + 2 \,cos\, x - 1] = 0$

$\left. \begin{array}{l}x = (2n - 1)\frac{\pi }{6} \Rightarrow \frac{\pi }{6},\frac{{3\pi }}{6},\frac{{5\pi }}{6} = 3\{2^{nd}} \,\,  equation\,\, gives \,\,cos{\mkern 1mu} x = \frac{{1 \pm \sqrt 2 }}{4} = 2\end{array} ight] = 5$

The number of solution of the equation,$\sum\limits_{r = 1}^5 {\cos (r\,x)} $ $= 0$ lying in $(0, \pi)$ is :

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