1、
f(x)=1/4(-cos2x-√3/2)+√3/2*[1-cos(2x-π/2)]/2
=-1/4cos2x-√3/8+√3/4-√3/4*sin2x
=-1/2*sin(2x+π/4)+√3/8
f(x)=√3/8
所以sin(2x+π/4)=0
2x+π/4=kπ
x∈{x|x=kπ/2+π/8,k∈Z}
2、
-π/6