f(x)=sin方x+2sinxcosx+3cos方x-2
=sin²x+cos²x+sin2x+2cos²x-2
=1+sin2x+1+cos2x-2
=sin2x+cos2x
=√2sin(2x+π/4)
1.函数的最小正周期 T=2π/2=π
2.由2kπ-π/2≤2x+π/4≤2kπ+π/2,k∈Z
得：kπ-3π/8≤x≤kπ+π/8,k∈Z
∴函数的单调递增区间是
[kπ-3π/8,kπ+π/8],k∈Z
由2kπ+π/2≤2x+π/4≤2kπ+3π/2,k∈Z
得：kπ+π/8≤x≤kπ+5π/8,k∈Z
∴函数的单调递减区间是
[kπ+π/8,kπ+5π/8],k∈Z
3
当2x+π/4=2kπ+π/2,即x=kπ+π/8,k∈Z时
f(x)取得最大值√2,此时x集合为
{x|x=kπ+π/8,k∈Z}
当2x+π/4=2kπ-π/2,即x=kπ-3π/8,k∈Z时
f(x)取得最大值-√2,此时x集合为
{x|x=kπ-3π/8,k∈Z}