已知函数f(x)=cos^4(x)+（2根号3）sinxcosx-sin^4(x)

f(x)=cos2x+(√3)sin2x=2cos(2x-π/3) 故Tmin=2π/2=π 单增区间：由-π+2kπ≤2x-π/3≤2kπ,-2π/3+2kπ≤2x≤2kπ+π/3,得-π/3+kπ≤x≤kπ+π/6 当x∈[0,π/2]时,f(x)∈[-1,2]