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Q.

The least positive value of x satisfying

sin2⁡2x+4sin4⁡x−4sin2⁡xcos2⁡x4−sin2⁡2x−4sin2⁡x=19 is

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a

2π/3

b

5π/6 

c

π/3

d

π/6

answer is B.

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Detailed Solution

4sin4⁡x4cos2⁡x−4sin2⁡xcos2⁡x=19⇒sin4⁡xcos4⁡x=19⇒tan4⁡x=19⇒tan⁡x=13⇒x=π/6

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