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The equation `2sin^3theta+(2lambda-3)sin^2theta-(3lambda+2)sintheta-2lambda=0` has exactly three roots in `(0,2pi)` , then `lambda` can be equal to 0

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The equation `2sin^3theta+(2lambda-3)sin^2theta-(3lambda+2)sintheta-2lambda=0` has exactly three roots in `(0,2pi)` , then `lambda` can be equal to 0 (b) 2 (c) 1 (d) `-1`
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Correct Answer - A::C::D
The equation becomes
`(sin theta -2) (sin theta + lambda) (2 sin theta+1)=0`
`rArr lambda= pm 1, 0`.
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