求函数f(x)=2sin^2x +cos(x +π/3）单调增区间

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
f(x)=1/2cos 2x+√3/2sin 2x+2(sin x-cos x)(xin x+cos x)
=1/2cos 2x+√3/2sin 2x-2(cos^2 x-sin^2 x)
=1/2cos 2x+√3/2sin 2x-2cos 2x
=-3/2cos 2x+√3/2sin 2x
=√3(1/2sin 2x-√3/2cos 2x)
=√3sin(2x-π/3)
1.可见,f(x)最小正周期为2π/2=π
图像为g(x)=√3sin(2x)向右平移π/6个单位得到
则对称轴为,x=kπ/2+5π/12(k∈Z)
2.令z=2x-π/3
由x∈[-π/12,π/2]
则z∈[-π/2,2π/3]
√3sin z∈[-√3,√3]
即f(x)值域为[-√3,√3]

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
f(x)=1/2cos 2x+√3/2sin 2x+2(sin x-cos x)(xin x+cos x)
=1/2cos 2x+√3/2sin 2x-2(cos^2 x-sin^2 x)
=1/2cos 2x+√3/2sin 2x-2cos 2x
=-3/2cos 2x+√3/2sin 2x
=√3(1/2sin 2x-√3/2cos 2x)
=√3sin(2x-π/3)