Question 750992
After expanding and simplifying we can write

{{{
(sin2x+cos2x)^2
}}}


as


{{{sin4x+1
}}}


Therfore

{{{
(sin2x+cos2x)^2=1
}}}


becomes 


{{{sin4x+1=1
}}}


and so



{{{sin4x=0
}}}



which has solutions...


{{{x=(n*pi)/4}}}


:)