y=sin(2x+π/3)sin(π/6-2x)的最小正周期怎么判断?y=sin(2x+π/3)sin(π/6-2x)的最小正周期怎么判断?是化简?用2π/w这种类型题怎么办?

Question

y=sin(2x+π/3)sin(π/6-2x)的最小正周期怎么判断?
y=sin(2x+π/3)sin(π/6-2x)的最小正周期怎么判断?
是化简?用2π/w
这种类型题怎么办?

Answer 1

sin（2x+π/3）=sin(2X+π/2-π/6)=cos(2x-π/6)
所以 原式=1/2(sin(π/3-4x))
所以 T=2π/欧米伽=π/2