已知函数f(x)=sin^2x+sinxcosx

题目

已知函数f(x)=sin^2x+sinxcosx
(1)求其最小正周期
(2)当0≤x≤π/2时,求其最值及相应的x值
(3)试求不等式f(x)≥1的解集

答案

f(x)=sin²x+sinxcosx
=[1-cos(2x)]/2 +sin(2x)/2
=sin(2x) /2 -cos(2x) /2 +1/2
=(√2/2)sin(2x-π/4)+1/2
最小正周期T=2π/2=π
0≤x≤π/2 -π/4≤2x-π/4≤3π/4 -√2/2≤sin(2x-π/4)≤1
sin(2x-π/4)=1时,f(x)有最大值[f(x)]max=(√2+1)/2
sin(2x-π/4)=-√2/2时,f(x)有最小值[f(x)]min=0
f(x)≥1
(√2/2)sin(2x-π/4)+1/2≥1
sin(2x-π/4)≥√2/2
2kπ+π/4≤2x-π/4≤2kπ+3π/4 (k∈Z)
kπ+π/4≤x≤kπ+π/2 (k∈Z)
x的解集为[kπ+π/4,kπ+π/2] (k∈Z)