y=sin^4 x + cos²x
=sin^4 x - sin²x +1
=(sin²x -1/2)² +3/4
=[(1-cos2x)/2 -1/2]² +3/4
=cos²2x /4 +3/4
=[(1+cos4x)/2] /4 +3/4
= cos4x / 8 +7/8
则最小正周期为 2π/4=
π/2.
y=sin^4 x + cos²x
=sin^4 x - sin²x +1
=(sin²x -1/2)² +3/4
=[(1-cos2x)/2 -1/2]² +3/4
=cos²2x /4 +3/4
=[(1+cos4x)/2] /4 +3/4
= cos4x / 8 +7/8
则最小正周期为 2π/4=
π/2.