(1 + cos4x) sin2x >= cos^2 2x
2cos^2 2x × sin2x - cos^2 2x >=0
cos^2 2x(2sin2x - 1)>= 0
cos^2 2x >= 0, x € R
2sin2x >= 1
sin2x >= 1/2
π/6 + 2πk =< 2x =< 5π/6 + 2πk
π/12 + πk =< x =< 5π/12 + πk ; k € Z
x € [π/12 + πk ; 5π/12 + πk] ; k € Z
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Answers & Comments
(1 + cos4x) sin2x >= cos^2 2x
2cos^2 2x × sin2x - cos^2 2x >=0
cos^2 2x(2sin2x - 1)>= 0
cos^2 2x >= 0, x € R
2sin2x >= 1
sin2x >= 1/2
π/6 + 2πk =< 2x =< 5π/6 + 2πk
π/12 + πk =< x =< 5π/12 + πk ; k € Z
x € [π/12 + πk ; 5π/12 + πk] ; k € Z