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| 1. |
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 to0 (b)2 (c) 1(d) `-1`A. -1B. 0C. `(1)/(2)`D. 1 |
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