f(x)=根号3sin2x+2sin(π/4+x)sin(π/4-x)

f(x)=根号3sin2x+2sin[π/2-(π/4-x)]sin(π/4-x)
=根号3sin2x+2cos(π/4-x)sin(π/4-x)
=根号3sin2x+sin(π/2-2x)
=根号3sin2x+cos2x
=2(根号3/2sin2x+1/2cos2x)
=2(cosπ/6sin2x+sinπ/6cos2x)
=2sin(2x+π/6)
最小正周期 T=2π/2=π
2kπ-π/2≤2x+π/6≤2kπ+π/2
2kπ-2π/3≤2x≤2kπ+π/3
kπ-π/3≤x≤kπ+π/6 (k∈Z)
∴单调递增区间为 [kπ-π/3 ,kπ+π/6 ] (k∈Z)
当x∈(0,π/2]
∴2x+π/6 ∈ (π/6 , 7π/6]
f(x)max=f(π/6)=2
f(x)min=f(0)=1
∴当 x∈(0,π/2] 时
值域为（0 , 2]
好辛苦啊! 慌忙中不知道有没有错误,若有的话望见谅哈!

f(x)=√3*Sin2x+2sin(π/4+x)sin(x-π/4)【Sinf(x)是偶函数】
=√3*Sin2x-2sin(π/4+x)sin(x+π/4)【Sinf(x)=-Sinf(x+π/2)】
=√3*Sin2x+Sin2x【Sinf(x)=-Sinf(x-π/2)】
=(√3+1)Sin2x
(1)π
(2)[-π/4+kπ,π/4+kπ](k...