设函数f(x)=sin2x+2cos^2x+1
3个回答

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+2

2x=2kπ+π/4

x=kπ+π/8

即最大值√2+2,此时x=kπ+π/8 (k∈Z)

(2)

最小正周期T=2π/w=2π/2=π