The number of solutions of the equation cos 6 | vedclass

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

Question

(D) Given the equation: $\cos 6x + \cos 4x + \cos 2x = -1$ in $[0, \pi]$.Rearranging the terms: $(\cos 6x + 1) + (\cos 4x + \cos 2x) = 0$.Using the id

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Given the equation: $\cos 6x + \cos 4x + \cos 2x = -1$ in $[0, \pi]$.Rearranging the terms: $(\cos 6x + 1) + (\cos 4x + \cos 2x) = 0$.Using the identity $\cos 2	heta = 2\cos^2 	heta - 1$,we have $1 + \cos 6x = 2\cos^2 3x$.Using the sum-to-product formula $\cos C + \cos D = 2\cos(\frac{C+D}{2})\cos(\frac{C-D}{2})$,we have $\cos 4x + \cos 2x = 2\cos 3x \cos x$.Substituting these into the equation: $2\cos^2 3x + 2\cos 3x \cos x = 0$.Factoring out $2\cos 3x$: $2\cos 3x(\cos 3x + \cos x) = 0$.Using the sum-to-product formula again: $2\cos 3x(2\cos 2x \cos x) = 0$,which simplifies to $4\cos x \cos 2x \cos 3x = 0$.This implies $\cos x = 0$ or $\cos 2x = 0$ or $\cos 3x = 0$.For $\cos x = 0$ in $[0, \pi]$,$x = \frac{\pi}{2} (90^{\circ})$.For $\cos 2x = 0$ in $[0, \pi]$,$2x = \frac{\pi}{2}, \frac{3\pi}{2} \implies x = \frac{\pi}{4} (45^{\circ}), \frac{3\pi}{4} (135^{\circ})$.For $\cos 3x = 0$ in $[0, \pi]$,$3x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2} \implies x = \frac{\pi}{6} (30^{\circ}), \frac{\pi}{2} (90^{\circ}), \frac{5\pi}{6} (150^{\circ})$.The distinct solutions are $\{30^{\circ}, 45^{\circ}, 90^{\circ}, 135^{\circ}, 150^{\circ}\}$.Thus,there are $5$ solutions.

The number of solutions of the equation $\cos 6x + \cos 4x + \cos 2x = -1$ in the interval $[0, \pi]$ is:

Similar Questions

The number of solutions of the trigonometric equation $1+\cos x \cdot \cos 5 x=\sin ^2 x$ in $[0, 2 \pi]$ is

The set of values of $x$ satisfying the equation $2^{\tan(x - \pi/4)} - 2(0.25)^{\frac{\sin^2(x - \pi/4)}{\cos 2x}} + 1 = 0$ is:

If $\alpha$ is a root of the equation $25 \cos^2 \theta + 5 \cos \theta - 12 = 0$,for $\frac{\pi}{2} < \alpha < \pi$,then $\sin 2\alpha =$

The number of values of $x$ in the interval $[0, 3\pi]$ satisfying the equation $2 \sin^2 x + 5 \sin x - 3 = 0$ is

The general solution of $\tan 3x = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module