数学卷22：函数f(x)=sin2ωx-2sin²ωx(ω>0)(1)若函数f(x)的周期为π,求ω的值；

f(x)=sin2ωx-2sin²ωx(ω>0)
=sin2ωx+1-2sin²ωx-1
=sin2ωx+cos2ωx-1
=√2sin(2ωx+π/4)-1
f(x)的周期为π
2π/2ω=π
ω=1
(2)
ω>0
f(x)=√2sin(2ωx+π/4)-1
令-π/2+2kπ