The number of solutions of the equation sum_{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$5$

Vedclass · Answer

(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:

Similar Questions

If $4(\sin 2x \sin 4x + \sin^2 x) = 3$,then $x =$

The number of values of $x$ for which $\sin 2x + \cos 4x = 2$ is

The number of solutions of the equation $4 \cos 2 \theta \cdot \cos 3 \theta = \sec \theta$,when $0 < \theta < \pi$,is

If $\tan (A - B) = 1$ and $\sec (A + B) = \frac{2}{\sqrt{3}}$,then the smallest positive value of $B$ is

If $\cos 7\theta = \cos \theta - \sin 4\theta$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module