17,已知函数f(x)=sin^2ωx+√3cosωx(ω>0)的最小正周期为π (1)求ω的值及函数f

17、已知函数f(x)=sin^2ωx+√3cosωx(ω＞0)的最小正周期为π
（1）求ω的值及函数f(x)的单调增区间
（2）求函数f(x)在[0,2π/3]上的值域
(1)解析：∵函数f(x)=sin^2ωx+√3cosωx(ω＞0)的最小正周期为π
Ω=2π/π=2
∴f(x)=(sin2x)^2+√3cos2x
令f’(x)=4sin2xcos2x-2√3sin2x=2sin2x(2cos2x-√3)=0
sin2x=0==>2x=2kπ==>x1=kπ,2x=2kπ+π==>x2=kπ+π/2
2cos2x-√3=0==>2x=2kπ-π/6==>x3=kπ-π/12,2x=2kπ+π/6==>x4=kπ+π/12
f’’(x)=8cos4x-4√3cos2x==> f’’(x1)=8-4√3>0,f’’(x2)=8+4√3>0,f’’(x3)=f’’(x4)=4-6

f(x)=sin^2ωx+√3cosωx(ω＞0)
=2sin(2wx+π/3])
T=2π/2w=π,解得w=1
f(x)=2sin(2x+π/3)