The number of solutions of the equation cos 6 | vedclass

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(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

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$5$

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(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:

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