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.函數(shù)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
∴函數(shù)的單調(diào)遞增區(qū)間是
[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
∴函數(shù)的單調(diào)遞減區(qū)間是
[kπ+π/8,kπ+5π/8],k∈Z
3
當(dāng)2x+π/4=2kπ+π/2,即x=kπ+π/8,k∈Z時(shí)
f(x)取得最大值√2,此時(shí)x集合為
{x|x=kπ+π/8,k∈Z}
當(dāng)2x+π/4=2kπ-π/2,即x=kπ-3π/8,k∈Z時(shí)
f(x)取得最大值-√2,此時(shí)x集合為
{x|x=kπ-3π/8,k∈Z}