已知向量a=(cosx,-1/2),b=(根号3sinx,cos2x) x属于R 设f(x)=a*b 求f(x)最小正周期 f(x)在[0,2派]上的最大值和最小值

f(x)=a*b    =√3cosxsinx-1/2cos2x    =√3/2sin2x-1/2cos2x    =sin(2x-π/6)最小正周期为；T=2π/2=π2.∵x∈[0,π/2]∴2x-π/6∈[-π/6,5π/6]∴当2x-π/6=-π/6时f(x)取得最小值...