The number of solutions of the equation sum_{ | vedclass

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(C) The given equation is $\cos(x) + \cos(2x) + \cos(3x) + \cos(4x) + \cos(5x) = 0$. Using the sum formula for cosines in arithmetic progression: $\su

Vedclass · Accepted Answer

$5$

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(C) The given equation is $\cos(x) + \cos(2x) + \cos(3x) + \cos(4x) + \cos(5x) = 0$. Using the sum formula for cosines in arithmetic progression: $\sum_{r=1}^{n} \cos(rx) = \frac{\sin(nx/2)}{\sin(x/2)} \cos\left(\frac{(n+1)x}{2}ight)$. For $n=5$,we have $\frac{\sin(5x/2)}{\sin(x/2)} \cos(3x) = 0$. This implies $\sin(5x/2) = 0$ or $\cos(3x) = 0$,provided $\sin(x/2) 
eq 0$. Case $1$: $\cos(3x) = 0$. In $(0, \pi)$,$3x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2} \implies x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$. Case $2$: $\sin(5x/2) = 0$. In $(0, \pi)$,$5x/2 = \pi, 2\pi, 3\pi, 4\pi \implies x = \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5} (	ext{outside}), \dots$. Checking the condition $\sin(x/2) 
eq 0$ (which is true for $x \in (0, \pi)$) and ensuring no overlap,the solutions are $x = \frac{\pi}{6}, \frac{2\pi}{5}, \frac{\pi}{2}, \frac{4\pi}{5}, \frac{5\pi}{6}$. There are $5$ solutions in total.

The number of solutions of the equation $\sum_{r=1}^{5} \cos(rx) = 0$ lying in the interval $(0, \pi)$ is:

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