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Question

If sin24x+cos2x=2sin4x.cos4x then number of values of x satisfying, if x[2π,2π] is


A

0

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B

2

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C

3

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D

4

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Solution

The correct option is A

0


sin24x+cos2x=2sin4xcos4xsin24x2sin4x cos4x+cos2xsin4x=2cos4x±4cos8x4cos2x2sin4x=cos4x±cos2x(cos6x1)cos6x=1 sin4x=1 4x=(4n2+1)π2cosx=±1 x=(4n2+1)π8x=n1π,n,Z n1π=(4n2+1)π8
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