(1)
f(x)=2✔3sinxcosx-2cos^2x+1
=✔3sin2x-cos2x
=2[✔3/2sin2x-1/2cos2x]
=2(sin2xcosπ/6-cos2xsinπ/6)
=2sin(2x-π/6)
T=2π/2=π
增区间:2kπ-π/2≤2x-π/6≤2kπ+π/2
2kπ-π/3≤2x≤2kπ+2π/3
kπ-π/6≤x≤kπ+π/3
即【kπ-π/6,kπ+π/3】
(2)f(x) =2sin(2x-π/6)
f(A/2) =2sin(A-π/6)=2
sin(A-π/6)=1
A-π/6=π/2
A=2π/3
a平方=b平方+c平方-2bccosA
=1+4-2×(-1/2)
=6
a=√6