设函数f（x）=（sinωx+cosωx）^ 2+2cos^2 ωx（ω＞0）的最小正周期为2π/3

（1）f（x）=sin²ωx+cos²ωx+2sinωxcosωx+2cos²ωx=1+2sinωxcosωx+2cos²ωx
=2+2sinωxcosωx+（2cos²ωx－1）=sin2ωx+cos2ωx+2=√2sin（2ωx+π/4)+2
最小正周期T=2π/2ω=π/ω=2π/3 ,所以ω=3/2,f（x）=√2sin（3x+π/4)+2
（2）将f（x）的图像向右平移π/2个单位长度即得y=g（x）,
所以g（x）=f（x－π/2）=√2sin（3x－3π/2+π/4)=√2sin（3x－5π/4)+2
当－π/2+2kπ ≤ 3x－5π/4 ≤ π/2+2kπ时g（x）单调递增
解不等式得π/4 +2kπ/3 ≤ x ≤ 7π/12 +2kπ/3
则y=g（x）的单调增区间为[ π/4 +2kπ/3,7π/12 +2kπ/3 ]

f（x）=（sinωx+cosωx）^ 2+2cos^2 ωx=1+2sinωxcosωx+2cos^2ωx
=1+2cosωx(sinωx+cosωx)=1+2√2cosωxsin(ωx+π/4)
=1+√2[sin(2ωx+π/4)+sinπ/4]
=1+√2sin(2ωx+π/4)+1
=√2sin(2ωx+π/4)+2
最小正周期为2π/3 ,T=2...

1 fx=√2sin2ωx+π/4+2 所以ω=3/2
2 fx=√2sin3x+π/4+2 所以gx=√2sin[3x-π/2+π/4]+2=√2sin3x-5π/4+2 因为3x-5π/4在[2kπ-π/2,2kπ+π/2]k属于Z 上递增 所以gx在[2kπ/3+3π/12,2kπ/3+7π/12]k属于Z 上递增 因为3x-5π/4在[2kπ+π/2,2kπ+3π/2]k属于Z 上递减 所以3x-5π/4在[2kπ/3+7π...