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Question

The number of solutions of sin3xcosx+sin2xcos2x+sinxcos3x=1 in [0,2π] is-

A
4
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B
2
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C
1
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D
0
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Solution

The correct option is D 0
sinxcosx[sin2x+sinxcosx+cos2x]=1sinxcosx+(sinxcosx)2=1
sin22x+2sin2x4=0sin2x=2±4+162=1±5, which is not possible.
Hence (D) is the correct answer.

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