Question

已知函数f（x）=2sin^x+2sinxcosx-1,求F（x）取得最小值时x的集合

Answer 1

f(x)=sin2x-(1-2sin^x)
=sin2x-cos2x
=√2sin(2x-π/4)
当sin(2x-π/4)=-1，即2x-π/4=2kπ-π/2 x=kπ-π/8 (k∈Z)时
ymin=-√2

Answer 2

f(x)=2sin²x+2sinxcosx-1
f(x)=1-cos2x+sin2x-1
f(x)=sin2x-cos2x
f(x)=√2sin(2x-π/4)

f(x)取得最小值时,有2x-π/4=2Kπ-π/2
即集合是{X|X=Kπ-π/8 }

Answer 3

f(x)=1-cos2x+sin2x-1
f(x)=sin2x-cos2x
f(x)=2^1/2sin(2x-pai/4)
sin(2x-pai/4)=-1
2x-pai/4=2kpai-pai/2
2x=2kpai-pai/4
x=kpai-pai/8(k:Z)