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Question

Sum of the roots of 3sin2x+2cos2x+31sin2x+2sin2x=28 in [2π,2π] is


A

0

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B
π
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C
2π
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D
3π2
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Solution

The correct option is B π

3sin2x+2cos2x+31sin2x+2sin2x=283sin2x+2cos2x+31sin2x+2(1cos2x)=283sin2x+2cos2x+33.3sin2x2cos2x=28
Let t=3sin2x+2cos2xt+33t=28t228t+27=0t=1, 27sin2x+2cos2x=0 or sin2x+2cos2x=32 sinx cosx+2cos2x=0 i.e sin2x=1, 2cos2x=1 [which is not possible. ]
2cosx(sinx+cosx)=0cosx=0, sinx+cosx=0
cosx=0 x=π2,π2,3π2,3π2.sinx+cosx=0tanx=1x=π4,3π4,7π4,5π4sum=π


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