f（x）=sin²x+根号3sinxcosx+2cos²x

f(x)=sin²x+√3sinxcosx+2cos²x
=1+cos²x+(√3/2)sin2x
=1+(1+cos2x)/2+(√3/2)sin2x
=(√3/2)sin2x+(1/2)cos2x+3/2
=sin(2x+π/6)+3/2
最小正周期 T=2π/2=π
-π/2+2kπ≤2x+π/6≤π/2+2kπ
-π/3+kπ≤x≤π/6+kπ
单调递增区间为 [-π/3+kπ,π/6+kπ] k=0,1,2,.,..
f(x)=sin(2x+π/6)+3/2=sin[2(x+π/12)]+3/2
是由 y=sin2x 左移π/12 再上移 3/2得到的
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f（x）=sin²x+√3sinxcosx+2cos²x
=1+√3sinxcosx+cos²x
=1+√3/2sin2x+1/2cos2x+1/2
=3/2+sin2xcosπ/6+cos2xsinπ/6)
=3/2+sin(2x+π/6) ...