已知函数f(x)=跟号3sin2x+2cos²x-1

问题描述:

已知函数f(x)=跟号3sin2x+2cos²x-1
(1)求最小正周期
(2)若x∈「0,π/2」求最小值

f(x)=跟号3sin2x+2cos²x-1
=√3sin2x+cos2x
=2(√3/2*sin2x+1/2cos2x)
=2(cosπ/6sin2x+sinπ/6cos2x)
=2sin(2x+π/6)
∴最小正周期是π
(2)最小值 f(π/2)=sin(7π/6)=-1/2