数学三角函数已知f(x)=2cosxsin(x+π/3)-根号3sinx^2+sinxcosx（1）求fx最小正周期（2）求fx的单调增区间（3）当x∈[0,π/4]时,求fx的值域.

f(x)=2cosxsin(x+π/3)-根号3sinx^2+sinxcosx
=2cosxsin(x+π/3)-2sinx[根号3/2sinx^2-1/2cosx]
=2cosxsin(x+π/3)+2sinxcos(x+π/3)
=2sin(2x+π/3)
最小正周期：Tmin=2π/2=π
(2)
当2kπ-π/2≤2x+π/3≤2kπ+π/2 (k∈Z)时，函数单调递增。此时kπ-5π/12≤x≤kπ+π/12
当2kπ+π/2≤2x+π/3≤2kπ+3π/2 (k∈Z)时，函数单调递减。此时kπ+π/12≤x≤kπ+7π/12
(k∈Z)
(3)
x∈[0，π/4]，π/3≤2x+π/3≤5π/6
2x+π/3=5π/6时，有f(x)min=1
2x+π/3=π/2时，有f(x)max=2
函数的值域为[1，2]

(1)
f(x)=2cosxsin(x+π/3)-√3sin²x+sinxcosx
=2cosx[sinx/2+√3cosx/2]-√3sin²x+sinxcosx
=sinxcosx+√3cos²x-√3sin²x+sinxcosx
=2sinxcosx+√3(cos²x-sin²x)
=sin(2x)+√3cos(2x)
=2[(1/2)sin(2x)+(√3/2)cos(2x)]
=2sin(2x+π/3)
最小正周期：T=2π/2=π
(2)
当2kπ-π/2≤2x+π/3≤2kπ+π/2 (k∈Z)时，单调递减，此时kπ-5π/12≤x≤kπ+π/12
当2kπ+π/2≤2x+π/3≤2kπ+3π/2 (k∈Z)时，单调递增，此时kπ+π/12≤x≤kπ+7π/12
单调增区间为：[kπ-5π/12，kπ+π/12]；单调递减区间为:[kπ+π/12，kπ+7π/12] (k∈Z)
(3)
x∈[0，π/4]，π/3≤2x+π/3≤5π/6
2x+π/3=5π/6时，有f(x)min=1
2x+π/3=π/2时，有f(x)max=2
所以：函数的值域为[1，2]

f(x)=cosxsinx+根号3cosx^2-根号3sinx^2+sinxcosx
=sin2x+根号3cos2x
=2sin（2x+π/3）
（1）T=2π/2=π
（2）2kπ-π/2 kπ-5π/12∴增区间(kπ-5π/12，kπ+π/12)，（3）x∈[0，π/4]
2x+π/3∈[π/3，5π/6]
∴f（x）∈[1，2]

化嘛，展开有，f(x)=2cosxsinx+根号3(cos²x-sin²x)=sin2x+根号3cos2x
=2（cos60sin2x+sin60cos2x）=2sin(2x+60)
应该可以做了

(1)f(x)=2cosxsin(x+π/3)-√3sin²x+sinxcosx=2cosx[sinx/2+√3cosx/2]-√3sin²x+sinxcosx=sinxcosx+√3cos²x-√3sin²x+sinxcosx=2sinxcosx+√3(cos²x-sin²x)=sin(2x)+√3cos(2x)=2[(...

f(x)=cosxsinx+根号3cosx^2-根号3sinx^2+sinxcosx
=sin2x+根号3cos2x
=2sin（2x+π/3）
（1）T=2π/2=π
（2）2kπ-π/2 kπ-5π/12∴增区间【kπ-5π/12，kπ+π/12】，k为整数
（3）x∈[0，π/4]
2x+π/3∈[π/3，5π/6]
∴f（x）∈【1，2】

先把能化开的都化开，然后把能配的用公式配好，化简就行了