已知函数f(x)=4sinxsin^2(π/4-x/2)-cos2x. 0分(1)、求f(2x-派/6)的单调增区间.(2)、若A是锐角△ABC的内角,f(A)=(根号3)-1,a=1,求△ABC面积的最大值.

f(x) = 4sinxsin^2(pi/4-x/2)-cos2x = 2sinx (1-cos(pi/2 - x))- cos2x = 2sinx (1-sinx) - cos2x
= 2sinx - 2sin^2 x - cos2x = 2sinx - (1-cos2x) - cos2x = 2sinx -1 + cos2x - cos2x = 2sinx - 1
f(2x-pi/6) = 2sin(2x-pi/6) - 1
when pi/2 >= 2x-pi/6 >= 3pi/2 or 5pi/6 f(A) = sqrt(3)-1 = 2sinA - 1, A = pi/3
a/sinA = b/sinB = c/sin(A+B) = 2/sqrt(3)
b = 2sinB/sqrt(3)
c = 2sin(A+B)/sqrt(3)
when A=B=C, △ABC面积的最大值 sqrt(3)/4

f(x)=4sinxsin^2(π/4-x/2)-cos2x
=2sinx[1-cos(π/2-x)]-cos2x
=2sinx-2sin^2x-cos2x
=2sinx-1
f(2x-派/6)=2sin(2x-π/6)
令2kπ-π/2