设函数f(x)=sin2x+2cos^2x+1

题目

设函数f(x)=sin2x+2cos^2x+1
求f(x)的最大值及相应x的取值
求f(x)的最小正周期

答案

f(x)=sin2x+2cos²x+1=sin2x+2cos²x-1+2=sin2x+cos2x+2=√2(sin2xcosπ/4+cos2xsinπ/4)+2=√2sin(2x+π/4)+2（1）-1≤sin(2x+π/4)≤1当sin(2x+π/4)=1即2x+π/4=2kπ+π/2时取得最大值√2+22x=2kπ+π/4x...