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

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Question

$3 \cos ^{2} 2 	heta+6 \cos 2 	heta-10 \cos ^{2} 	heta+5=0$

$3 \cos ^{2} 2 	heta+6 \cos 2 	heta-5(1+\cos 2 	heta)+5=0$

$3 \cos ^{2} 2 	he

Vedclass · Accepted Answer

$32$

Vedclass · Answer

$3 \cos ^{2} 2 	heta+6 \cos 2 	heta-10 \cos ^{2} 	heta+5=0$

$3 \cos ^{2} 2 	heta+6 \cos 2 	heta-5(1+\cos 2 	heta)+5=0$

$3 \cos ^{2} 2 	heta+\cos 2 	heta=0$

$\operatorname{Cos} 2 	heta=0$ OR $\cos 2 	heta=-1 / 3$

$	heta \in[-4 \pi, 4 \pi]$

$2 	heta=(2 n +1) \cdot \frac{\pi}{2}$

$	herefore \quad 	heta=\pm \pi / 4 . \pm 3 \pi / 4 \ldots \ldots \pm 15 \pi / 4$

Similarly $\cos 2 	heta=-1 / 3$ gives $16$ solution

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

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