解
f(x)=2cos^2x+2√3sinxcosx-1
=√3sin2x+cos2x
=2sin(2x+π/6)
∴最小正周期為：2π/2=π不懂追問在三角形ABC中，角ABC所對(duì)的邊分別是abcf(c/2)=2且c²=ab試判斷三角形ABC的形狀f(c/2)=2則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∴是等邊三角形不懂追問，有些地方有平方，不知道手機(jī)看不看得見看不見的f(c/2)=2則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∴是等邊三角形我用^2代替了