If 4^(sin2x+2cos^2 x) +4^(1-sin2x+2sin^2 x) =65 then sin2x + cos2x has the value equal to

we can get 65 by 64+1 so
either
1-sin2x+2sin²x = 0 or sin2x+2cos²x=0\
sin2x+cos2x = -1
sin(2x+pie/4) = -1
so 2x+pie/4 =npie+(-1)ⁿ 5pie/4 find x from here

If u assume the value of(sin2x+2cos^2 x) as 3 then the factor(1-sin2x+2sin^2 x) will have value 0.
Hence by taking value of(sin2x+2cos^2 x) as 3 simplify and find the value of sin2x+cos2x