已知函数f（x）=sin눀x＋2sinxconx＋3con눀x.1求函数f（x）的单调曾区间,2.

f(x)=sin2x+cos2x+2
=√2sin(2x+π/4)+2
（1） 2kπ-π/2<=2x+π/4<=2kπ+π/2
kπ-3π/8<=x<=kπ+π/8
增区间【kπ-3π/8，kπ+π/8】 k∈Z
（2）f(C)=√2sin(2C+π/4)+2=3
sin(2C+π/4)=√2/2 2C+π/4=3π/4 C=π/4
余弦定理
c^2=a^2+b^2-2ab*cosC
2=4+b^2-2√2b
b^2-2√2b+2=0
(b-√2)^2=0
b=√2

f(x)=sin²x＋2sinxcosx＋3cos²x
=2sinxcosx＋2cos²x+1
=sin2x＋cos2x+2
=√2sin(2x+π/4)+2
1. 2kπ-π/2≤2x+π/4≤2kπ+π/2
kπ-3π/8≤x≤kπ+π/8
2kπ+π/2≤2x+π/4≤2kπ+3π/2
kπ+π/8≤x≤kπ+5π/8
单增区间:[kπ-3π/8,kπ+π/8];单减区间:[kπ+π/8,kπ+5π/8]
2. ∵c∴0f(C)=√2sin(2C+π/4)+2=3
2C+π/4=3π/4
C=π/4
c²=b²+a²-2abcosC
b²-2√2b+2=0
(b-√2)²=0
b=√2

1-cos^2x+3cos^2x+2sinxcosx
=cos2x+2+sin2x
=根2（sin（2x+派/4））+2

-派/2+2k派＜2x+派/4＜派/2+2k派
[-3派/4+k派，派/4+k派]

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