1）f(x)=2cos^2x+cos(2x+π/3)-1
=cos2x+cos2xcosπ/3-sin2xsinπ/3
=cos2x+1/2cos2x-√3/2sin2x
=3/2cos2x-√3/2sin2x
=√3(cos2xcosπ/6-sin2xsinπ/6)
=√3cos(2x+π/6)
所以最小值周期T=2π/w=2π/2=π
因为当(2x+π/6)∈（2kπ-π,2kπ）时,f(x)单调递增
此时x∈（kπ-7π/12,kπ-π/12)
所以f(x)单调增区间为（kπ-7π/12,kπ-π/12）（k∈Z）
2）f(a)=√3cos(2a+π/6)=-3/2
则cos(2x+π/6)=-√3/2
即2x+π/6=kπ+5π/6
x=kπ/2+π/3 (k∈Z）
因为a为锐角
所以a=π/3