先化简,
f(x)=3-2sin^2ωx-2cos(ωx+π/2)cosωx
=2+cos2ωx+sin2ωx
=2+√2sin(2ωx+π/4)
由于函数过点(π/16,2+√2) ,
取x为π/16,则2+√2sin(ωπ/8+π/4)=2+√2
得出ωπ/8+π/4=π/2+2kπ,
又由于0
先化简,
f(x)=3-2sin^2ωx-2cos(ωx+π/2)cosωx
=2+cos2ωx+sin2ωx
=2+√2sin(2ωx+π/4)
由于函数过点(π/16,2+√2) ,
取x为π/16,则2+√2sin(ωπ/8+π/4)=2+√2
得出ωπ/8+π/4=π/2+2kπ,
又由于0