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]=1⇒sinxcosx+(sinxcosx)2=1 sin22x+2sin2x−4=0⇒sin2x=−2±√4+162=−1±√5, which is not possible. Hence (D) is the correct answer.