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Question

The least potive value of x satisfying sin22x+4sin4x4sin2xcos2x4sin22x4sin2x=19 is

A
π3
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B
π6
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C
2π3
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D
5π6
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Solution

The correct option is A π6

We have,

sin22x+4sin4x4sin2xcos2x4sin22x4sin2x=19

sin22x+4sin4x(2sinxcosx)24(2sinxcosx)24sin2x=19

sin22x+4sin4xsin22x44sin2xcos2x4sin2x=19

4sin4x4(1sin2xcos2xsin2x)=19

sin4x1sin2xsin2xcos2x=19

sin4xcos2xsin2xcos2x=19

sin4xcos2x(1sin2x)=19

sin4xcos2xcos2x=19

sin4xcos4x=19

tan4x=19

tan4x=(13)4


On comparing both side and we get,

tan x=13

tanx=tanπ6


On comparing that,

x=π6


Hence, this is the answer.


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