函数y=sin2x+sin(2x+π/3)/cos2x +cos(2x+π/3)的最小正周期为

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函数y=sin2x+sin(2x+π/3)/cos2x +cos(2x+π/3)的最小正周期为

函数y=sin2x+sin(2x+π/3)/cos2x +cos(2x+π/3)的最小正周期为
函数y=sin2x+sin(2x+π/3)/cos2x +cos(2x+π/3)的最小正周期为

函数y=sin2x+sin(2x+π/3)/cos2x +cos(2x+π/3)的最小正周期为
3角函数题
这题首先声明是高中水平的
答案是π/2
y=sin2x+sin(2x+π/3)/cos2x +cos(2x+π/3)
=[(3/2sin2x)+(根3/2cos2x)]/[(3/2cos2x)-(根3/2sin2x)]
=sin(2x+π/6)/[cos(2x+π/6)]
=tan(2x+π/6)
TAN最小正周期为π
所以此题最小正周期为π/2