The equation 2sin^3theta+(2lambda-3)sin^...

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 (a)`0` (b) `2 `(c)` 1` (d) `-1`

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