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Questions  

If sin⁡α, sin⁡β are the roots of the equation asin2⁡θ+bsin⁡θ+c=0 and sin⁡α+2sin⁡β=1 and a2+2b2+3ab+ac=

a
– 1
b
0
c
1
d
a+b+c

detailed solution

Correct option is B

sin⁡α+sin⁡β=−b/a, sin⁡αsin⁡β=c/a, sin⁡α+2sin⁡β=1⇒sin⁡α=ca+b which satisfies the given equation (Note c≠0 ) and the required value is 0.

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