The least potive value of x satisfying sin22x+4sin4x−4sin2xcos2x4−sin22x−4sin2x=19 is
We have,
sin22x+4sin4x−4sin2xcos2x4−sin22x−4sin2x=19
⇒sin22x+4sin4x−(2sinxcosx)24−(2sinxcosx)2−4sin2x=19
⇒sin22x+4sin4x−sin22x4−4sin2xcos2x−4sin2x=19
⇒4sin4x4(1−sin2xcos2x−sin2x)=19
⇒sin4x1−sin2x−sin2xcos2x=19
⇒sin4xcos2x−sin2xcos2x=19
⇒sin4xcos2x(1−sin2x)=19
⇒sin4xcos2xcos2x=19
⇒sin4xcos4x=19
⇒tan4x=19
⇒tan4x=(1√3)4
On comparing both side and we get,
tan x=1√3
⇒tanx=tanπ6
On comparing that,
x=π6
Hence, this is the answer.