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
问题描述：
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
根据二倍角公式：cos2x=2cos²x-1
得2cos²x=cos2x+1
f(x)=sin2x+cos2x+1+1
=√2(sin2x·cos45º+cos2x·sin45º)+2
=√2sin(2x+45º)+2
其中ω＝2
最小正周期T=2π/ω=π
f(x)max=√2+2
f(x)min=2﹣√2
答
f(x)=sin2x+cos2x+2
=(√2)*sin(2x+π/4)+2
当2x+π/4=π/2 +2kπ 即：x=π/8 + kπ时 f(x)最大值是√2+2
周期T=2π/2=π
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+22x=2kπ+π/4x...