已知函数f(x)=acos^2x+bsinxcosx满足f(0)
问题描述:
已知函数f(x)=acos^2x+bsinxcosx满足f(0)
f(0)=f(圆周率/3)=2
(1)求函数f(x)的解析式和最小正周期T
(2)求函数f(x)的单调增区间
答
f(x)=acos^2x+bsinxcosx满足:f(0)=2,f(π/3)=1/2+√3/2,得a=2,1/2+b√3/4=1/2+√3/2,知b=2.从而f(x)=2cos^2x+2sinxcosx=cos2x+sin2x+1=√2sin(2x+π/4)+1.最小正周期T=2π/2=π单调增区间是:2kπ-π/2...