The number of elements in the set S = {theta | vedclass

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(A) Given equation: $3 \cos^2 2	heta + 6 \cos 2	heta - 10 \cos^2 	heta + 5 = 0$Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we g

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$32$

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(A) Given equation: $3 \cos^2 2	heta + 6 \cos 2	heta - 10 \cos^2 	heta + 5 = 0$Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:$3 \cos^2 2	heta + 6 \cos 2	heta - 10(\frac{1 + \cos 2	heta}{2}) + 5 = 0$$3 \cos^2 2	heta + 6 \cos 2	heta - 5 - 5 \cos 2	heta + 5 = 0$$3 \cos^2 2	heta + \cos 2	heta = 0$$\cos 2	heta(3 \cos 2	heta + 1) = 0$This gives two cases:Case $1$: $\cos 2	heta = 0$For $	heta \in [-4\pi, 4\pi]$,$2	heta \in [-8\pi, 8\pi]$.$\cos 2	heta = 0 \implies 2	heta = (2n+1)\frac{\pi}{2}$,where $n \in \{-8, -7, \dots, 7\}$.There are $16$ values for $2	heta$ in the interval $[-8\pi, 8\pi]$,so there are $16$ values for $	heta$.Case $2$: $\cos 2	heta = -\frac{1}{3}$Since $-1 In the interval $[-8\pi, 8\pi]$ (length $16\pi$),there are $8 	imes 2 = 16$ solutions.Total number of elements $= 16 + 16 = 32$.

The number of elements in the set $S = \{\theta \in [-4\pi, 4\pi] : 3 \cos^2 2\theta + 6 \cos 2\theta - 10 \cos^2 \theta + 5 = 0\}$ is

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