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

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(NONE) Given equation: $3 \cos^4 	heta - 5 \cos^2 	heta - 2 \sin^2 	heta + 2 = 0$.
Using $\sin^2 	heta = 1 - \cos^2 	heta$,we substitute:
$3 \c

Vedclass · Accepted Answer

$5$

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(NONE) Given equation: $3 \cos^4 	heta - 5 \cos^2 	heta - 2 \sin^2 	heta + 2 = 0$.
Using $\sin^2 	heta = 1 - \cos^2 	heta$,we substitute:
$3 \cos^4 	heta - 5 \cos^2 	heta - 2(1 - \cos^2 	heta) + 2 = 0$
$3 \cos^4 	heta - 5 \cos^2 	heta - 2 + 2 \cos^2 	heta + 2 = 0$
$3 \cos^4 	heta - 3 \cos^2 	heta = 0$
$3 \cos^2 	heta (\cos^2 	heta - 1) = 0$
$3 \cos^2 	heta (-\sin^2 	heta) = 0$
$-3 \cos^2 	heta \sin^2 	heta = 0$
This implies $\cos^2 	heta = 0$ or $\sin^2 	heta = 0$.
Case $1$: $\cos^2 	heta = 0 \implies \cos 	heta = 0$. In $[0, 2\pi]$,$	heta = \frac{\pi}{2}, \frac{3\pi}{2}$.
Case $2$: $\sin^2 	heta = 0 \implies \sin 	heta = 0$. In $[0, 2\pi]$,$	heta = 0, \pi, 2\pi$.
The set $S = \{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\}$.
The number of elements is $5$.

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

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