If S = {theta in [-pi, pi] : cos theta cos fr | vedclass

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(C) Given equation: $\cos 	heta \cos \frac{5	heta}{2} = \cos 7	heta \cos \frac{7	heta}{2}$.Multiply both sides by $2$: $2 \cos 	heta \cos \frac{5

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) Given equation: $\cos 	heta \cos \frac{5	heta}{2} = \cos 7	heta \cos \frac{7	heta}{2}$.Multiply both sides by $2$: $2 \cos 	heta \cos \frac{5	heta}{2} = 2 \cos 7	heta \cos \frac{7	heta}{2}$.Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we get:$\cos(\frac{7	heta}{2}) + \cos(-\frac{3	heta}{2}) = \cos(\frac{21	heta}{2}) + \cos(\frac{7	heta}{2})$.This simplifies to $\cos(\frac{3	heta}{2}) = \cos(\frac{21	heta}{2})$.General solution: $\frac{21	heta}{2} = 2n\pi \pm \frac{3	heta}{2}$.Case $1$: $\frac{21	heta}{2} - \frac{3	heta}{2} = 2n\pi \implies 9	heta = 2n\pi \implies 	heta = \frac{2n\pi}{9}$. For $	heta \in [-\pi, \pi]$,$n \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$ ($9$ values).Case $2$: $\frac{21	heta}{2} + \frac{3	heta}{2} = 2n\pi \implies 12	heta = 2n\pi \implies 	heta = \frac{n\pi}{6}$. For $	heta \in [-\pi, \pi]$,$n \in \{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$ ($13$ values).Combining both sets and removing duplicates: $	heta \in \{0, \pm \frac{\pi}{6}, \pm \frac{2\pi}{9}, \pm \frac{\pi}{3}, \pm \frac{4\pi}{9}, \pm \frac{\pi}{2}, \pm \frac{2\pi}{3}, \pm \frac{8\pi}{9}, \pm \pi\}$.Total distinct values $n(S) = 13$.

If $S = \{\theta \in [-\pi, \pi] : \cos \theta \cos \frac{5\theta}{2} = \cos 7\theta \cos \frac{7\theta}{2}\}$,then $n(S)$ is equal to . . . . . . .

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