y=(3/2)cos(1/2x-π/6)X∈R的周期 ,

问题描述:

y=(3/2)cos(1/2x-π/6)X∈R的周期 ,

在函数y = (3/2)cos(1/2 x -π/6)中,ω = 1/2
所以周期T = 2π/2 = 2π/(1/2) = 4π