Question

已知f(x)=sin²x+2sinxcosx+3cos²x，x∈(0,兀)求；1函数的最小正周期和值域2，求函数的单调递增区

Answer 1

y=1+sin2x+cos²x
=1+sin2x+1+cos2x
=2+sin2x+cos2x
=2+√2sin(2x+π/4)
所以周期为T=2π /2=π

(2) -π/2+2kπ<=2x+π/4<=π/2+2kπ
-3π/4+2kπ<=2x<=π/4+2kπ
-3π/8+kπ<=x<=π/8+kπ

Answer 2

f(x)=sin²x+2sinxcosx+3cos²x
=sin²x+cos²x+2sinxcosx +2cos²x
=1+2sinxcosx +2cos²x
=2+sin2x+2cos²x-1
=2+sin2x+cos2x
=2+√2(√2/2sin2x+√2/2cos2x)
=2+√2(sin2xcosπ/4+sinπ/4cos2x)

=√2sin(2x+π/4)+2
函数的最小正周期π
值域[2-√2,2+√2]
单调递增(-3π/8+kπ, π/8+kπ)

Answer 3

1f(x)=1+cos2x+1+sin2x=√2sin(2x+π/4)+2 T=π
∵x∈(0,π）∴（2x+π/4）∈(π/4,9π/4)
值域为[-2+√2，2+√2]
2 （2x+π/4）∈[ -π/2+2kπ,π/2+2kπ]
x∈[-π/8+kπ，π/8+kπ]即增区间。