Question

求解这道题目第一问，谢谢？

Answer 1

f(x)=sinx + sinxcos(π/6) + cosxsin(π/6) - cos[π+(x + π/3)]
=sinx + (√3/2)sinx + (1/2)cosx - [-cos(x + π/3)]
=sinx + (√3/2)sinx + (1/2)cosx + cos(x + π/3)
=sinx + (√3/2)sinx + (1/2)cosx + cosxcos(π/3) - sinxsin(π/3)
=sinx + (√3/2)sinx + (1/2)cosx + (1/2)cosx - (√3/2)sinx
=sinx + cosx
=√2sin(x + π/4)
∴T=2π/1=2π
单调递减区间是：
2kπ + π/2≤x + π/4≤2kπ + 3π/2
∴2kπ + π/4≤x≤2kπ + 5π/4，(k∈Z)