If cos 6theta + cos 4theta + cos 2theta + 1 = | vedclass

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Question

(D) Given equation: $\cos 6	heta + \cos 4	heta + \cos 2	heta + 1 = 0$ Group the terms: $(\cos 6	heta + \cos 2	heta) + (\cos 4	heta + 1) = 0$ Usi

Vedclass · Accepted Answer

$30^\circ, 45^\circ, 90^\circ, 135^\circ, 150^\circ$

Vedclass · Answer

(D) Given equation: $\cos 6	heta + \cos 4	heta + \cos 2	heta + 1 = 0$ Group the terms: $(\cos 6	heta + \cos 2	heta) + (\cos 4	heta + 1) = 0$ Using the identities $\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$ and $\cos 2A = 2\cos^2 A - 1$: $2\cos 4	heta \cos 2	heta + 2\cos^2 2	heta = 0$ $2\cos 2	heta (\cos 4	heta + \cos 2	heta) = 0$ $2\cos 2	heta (2\cos 3	heta \cos 	heta) = 0$ $4\cos 3	heta \cos 2	heta \cos 	heta = 0$ This implies $\cos 3	heta = 0$ or $\cos 2	heta = 0$ or $\cos 	heta = 0$. For $0 $1$) $\cos 3	heta = 0$ $\Rightarrow 3	heta = 90^\circ, 270^\circ, 450^\circ$ $\Rightarrow 	heta = 30^\circ, 90^\circ, 150^\circ$. $2$) $\cos 2	heta = 0$ $\Rightarrow 2	heta = 90^\circ, 270^\circ$ $\Rightarrow 	heta = 45^\circ, 135^\circ$. $3$) $\cos 	heta = 0 \Rightarrow 	heta = 90^\circ$. Combining these,the values are $	heta = 30^\circ, 45^\circ, 90^\circ, 135^\circ, 150^\circ$.

If $\cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 0$,where $0 < \theta < 180^\circ$,then $\theta = $

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