f(x)=2cosxcos(2π/3)+2sinxsin(2π/3)+2cosx
=√3sinx+cosx
=2sin(x+π/6)
∵ x∈[π/2,π]
则 x+π/6∈[2π/3,7π/6]
∴ sin(x+π/6)∈[-1/2,√3/2]
即2sin(x+π/6)∈[-1,√3]f(x)=2cosxcos(2π/3)+2sinxsin(2π/3)+2cosx=√3sinx+cosx 这一步怎么变化过来的没懂。展开化简就行