Question

已知函数f(x)=2sinx(sinx+cosx) 求使f(x)大于等于2成立的x的取值范围

Answer 1

f(x)=2sinx(sinx+cosx)=2(sinx)^2+2sinxcosx=sin2x-cos2x+1=√2sin(2x-π/4)+1。
f(x)>=2，则sin(2x-π/4)>=√2/2，2kπ+π/4<=2x-π/4<=2kπ+3π/4，即kπ+π/4<=x<=kπ+π/2。
所以，x的取值范围是[kπ+π/4,kπ+π/2]，k为整数。

Answer 2

f(x)=2sinx(sinx+cosx)
=2sin²x+2sinxcosx
=1-cos2x+sin2x
=1+√2sin（2x-π/4）
因为f(x)≥2成立，即1+√2sin（2x-π/4）≥2
√2sin（2x-π/4）≥1
sin（2x-π/4）≥√2/2
所以π/4+2kπ≤2x-π/4≤3π/4+2kπ
π/2+2kπ≤2x≤π+2kπ
π/4+kπ≤x≤π/2+kπ k∈z

Answer 3

f(x)=2sinx(sinx+cosx)
=2sin²x+2sinxcosx
=1-cos2x+sin2x
f(x)≥2
则：
sin2x-cos2x≥1
√2(sin2xcosπ/4-cos2xsinπ/4)≥1
sin(2x-π/4)≥√2/2
2kπ+π/4≤2x-π/4≤2kπ+3π/4
2kπ+π/2≤2x≤2kπ+π
kπ+π/4≤x≤kπ+π/2；k∈Z