已知函数fx =sinxsinx +2sinxcosx+3COSXCOS 求最大值及X的取值范围 单调增区间

f(x)=sinxsinx+2sinxcosx+3cosxcosx
f(x)=(sinx)^2+sin(2x)+3(cosx)^2
f(x)=1+sin(2x)+2(cosx)^2
f(x)=1+sin(2x)+1+cos(2x)
f(x)=2+sin(2x)+cos(2x)
f'(x)=2[cos(2x)-sin(2x)]
令：f'(x)=0,即：2[cos(2x)-sin(2x)]=0
整理,有：cos(2x)=sin(2x)
tan(2x)=1
有：2x=2kπ+π/4、2x=2kπ+5π/4
解得：x=kπ+π/8、x=kπ+5π/8
此时,f(x)有最大值：f(x)最大=2+sin(2×π/8)+cos(2×π/8)=2+[(√2)/2]^2=2+1/2=5/2
令：f'(x)＞0,即：2[cos(2x)-sin(2x)]＞0
整理,有：cos(2x)＞sin(2x)
1、当cos(2x)＞0时,有：tan(2x)＜1
解得：kπ-π/8＜x＜kπ+π/8、kπ+3π/8＜x＜kπ+5π/8
考虑到cos(2x)＞0,有：kπ-π/4＜x＜kπ+π/4
f(x)的单调增区间是：x∈(kπ-π/8,kπ+π/8)
2、当cos(2x)＜0时,有：tan(2x)＞1
解得：kπ+π/8＜x＜kπ+π/4、kπ+5π/8＜x＜kπ+3π/4
考虑到cos(2x)＜0,有：kπ+π/4＜x＜kπ+3π/4
f(x)的单调增区间是：x∈(kπ+5π/8,kπ+3π/4)
综上所述,f(x)的单调增区间是：x∈(kπ-π/8,kπ+π/8)∪(kπ+5π/8,kπ+3π/4)

f'(x)=2sinxcosx+2cosxcosx-2sinxsinx-6sinxcosx=2cosxcosx-2sinxsinx-4sinxcosx=2cos2x-2sin2x
令f'(x)>0,则cos2x>sin2x,根据图像可知-3π/8+kπ令f'(x)<0,π/8+kπf(max)=f(π/8)=2+根号2

f(x)=1+sin2x+cos2x-1
=(2)^1/2sin(x+PI/4)
max=2^(1/2)　　x属于实数
增区间：（2k*PI+PI/4,5*PI/4+2k*IP）
其中PI是圆周率