(1)
f(x)=sin(x-π/6)+cos(x-π/3),g(x)=2sin²x/2
f(x)=sinxcosπ/6-cosxsinπ/6+cosxcosπ/3+sinxsinπ/3
=√3sinx
∴f(α)=√3sinα=3√3/5
∴sinα=3/5
∵α是第一象限角
∴cosα=4/5
∴sin²α/2=(1-cosα)/2=1/5
∴g(α)=2sin²α/2=2/5
(2)
f(x)≥g(x)即√3sinx≥2sin²x/2
即√3sinx≥1-cosx
∴√sinx+cosx≥1
兩邊同時除以2
√3/2sinx+1/2cosx≥1/2
即sin(x+π/6)≥1/2
∴2kπ+π/6≤x+π/6≤2kπ+5π/6
∴x集合為{x}2kπ≤x≤2kπ+2π/3,k∈Z}