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 equation $3 \cos x + 4 \sin x = 6$ has

If $\tan^2 \theta - (1 + \sqrt{3}) \tan \theta + \sqrt{3} = 0$,then the general value of $\theta$ is

The general solution satisfying both the equations $\sin x = -\frac{3}{5}$ and $\cos x = -\frac{4}{5}$ is

If $\alpha, \beta, \gamma$ and $\delta$ are the solutions of the equation $\tan \left( \theta + \frac{\pi}{4} \right) = 3 \tan 3\theta$,no two of which have equal tangents,then the value of $\tan \alpha + \tan \beta + \tan \gamma + \tan \delta$ is

If $\cos \theta = -\frac{1}{\sqrt{2}}$ and $\tan \theta = 1$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module