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Let S = { theta in [-2pi,2pi]:2 cos^(2)...

Let `S = { theta in [-2pi,2pi]:2 cos^(2) theta + 3 sin theta = 0}`, then the sum of the equations elements of S is .

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The correct Answer is:
2
`2 cos^(2) theta+3 sin theta=0 rArr 2-2sin^(2)theta+3 sin theta=0`
`rarr 2 sin^(2) theta+3 sin theta- 2 =0 rArr 2 sin^(2)theta-4 sin^(2) theta+sin theta-2=0`
`(2sin theta + 1) ( sin theta-2) = 0`
`rArr sin theta= -1//2 rArr theta= (7pi)/(6), (11 pi)/(6), (-5pi)/(6), (-pi)/(6) rArr (7pi+11 pi - 5 pi - pi)/(6) = (12pi)/(6)=2pi`
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