Question

已知函数y（x）=sin（2x+π/6）+2/3,x∈R.（1）求函数f（x）的最小正周期和单调增区间； （2）函数f（

Answer 1

1

f(x)=sin[2x+(π/6)]+(2/3)

T=2π/ω=π

2kπ-(π/2)≤2x+(π/6)≤2kπ+(π/2)

单调递增区间{x|kπ-(π/3)≤x≤kπ+(π/6)}

2

向左平移π/12,向上平移2/3 (两步骤可以调换顺序)