Question

已知函数f（x）=sin²x+（根号下）3sinxcosx+2cos²x求最值、周期和单调区间

Answer 1

f(x)=(1-cos2x)/2+(√3/2)sin2x+1+cos2x
f(x)=1/2+(cos2x)/2+(√3/2)sin2x+1
f(x)=3/2+sin(2x+π/6)
最值为+-5/2
周期2π/2=π
单增区间2kπ-π/2<=2x+π/6<=2kπ+π/2
kπ-π/6<=x<=kπ+π/6
单减区间kπ+π/6<=2x<=kπ+5π/6

Answer 2

f(x)=1+√3/2*sin2x+cosx^2=1+√3/2*sin2x+(cos2x+1)/2=1+√3/2*sin2x+1/2*cos2x+1/2 =√3/2*sin2x+1/2*cos2x+3/2=sin2xcos30°+cos2xsin30°+3/2=sin(2x+30°)+3/2
最大值1+3/2=5/2，最小值-1+3/2=1/2
T=2pai/2=pai
单增区间：kpai-pai/2<=2x+30°<=kpai+pai/2,求x
单间区间：kpai+pai/2<=2x+30°<=kpai+3pai/2,求x