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Question

If 3cosx2sinx, then the general solution of sin2xcos2x=2sin2x is

A
nπ+(1)nπ2,nZ
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B
nπ2,nZ
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C
(4n±1)π2,nZ
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D
(2n1)π,nZ
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Solution

The correct option is C (4n±1)π2,nZ
Given equation is
sin2xcos2x=2sin2x
1cos2x(2cos2x1)=22sinxcosx
cosx(2sinx3cosx)=0
cosx=0 (2sinx3cosx0)
x=4n±1) π2

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