f(x)=(1+cos2x)/2-√3sin2x/2+1
=cos(2x+π/3)+3/2
(1)2kπ-π≤2x+π/3≤2kπ,k∈Z
所以闹递增区间为[kπ-2π/3,kπ-π/6],k∈Z
(2)f(a)=cos(2a+π/3)+3/2=5/6,所以cos(2a+π/3)=-2/3
因为2a∈(2π/3,4π/3),所以2a+π/3∈(π,5π/3)
sin(2a+π/3)=-√5/3
sin2a=sin[(2a+π/3)-π/3]=(2√3-√5)/6
f(x)=(1+cos2x)/2-√3sin2x/2+1
=cos(2x+π/3)+3/2
(1)2kπ-π≤2x+π/3≤2kπ,k∈Z
所以闹递增区间为[kπ-2π/3,kπ-π/6],k∈Z
(2)f(a)=cos(2a+π/3)+3/2=5/6,所以cos(2a+π/3)=-2/3
因为2a∈(2π/3,4π/3),所以2a+π/3∈(π,5π/3)
sin(2a+π/3)=-√5/3
sin2a=sin[(2a+π/3)-π/3]=(2√3-√5)/6