解
f(x)=2cos^2x+2√3sinxcosx-1
=√3sin2x+cos2x
=2sin(2x+π/6)
∴最小正周期为：2π/2=π不懂追问在三角形ABC中，角ABC所对的边分别是abcf(c/2)=2且c²=ab试判断三角形ABC的形状f(c/2)=2则2sin(c+π/6)=2即sin(c+π/6)=1∴c+π/6=π/2+2kπ∴c=π/3+2kπ∵c∈(0.π)∴c=π/3由余弦c²=a²+b²-2abcosC即a²+b²-ab=ab∴a²+b²-2ab=0即(a-b)²=0∴a=b∴A=B=C=60∴是等边三角形不懂追问，有些地方有平方，不知道手机看不看得见看不见的f(c/2)=2则2sin(c+π/6)=2即sin(c+π/6)=1∴c+π/6=π/2+2kπ∴c=π/3+2kπ∵c∈(0.π)∴c=π/3由余弦c^2=a^2+b^2-2abcosC即a^2+b^2-ab=ab∴a^2+b^2-2ab=0即(a-b)²=0∴a=b∴A=B=C=60∴是等边三角形我用^2代替了