已知函数f(x)=sinx*cosx+根号3*cosx^2,(1)求f(x)的最小正周期,(2),求f(x)在区间[-π

先化简,再求值.
f(x)=（1/2）2sinx*cosx+√3*cosx^2
=1/2sin2x+√3/2（2cos^2-1)+√3/2
=1/2sin2x+√3/2cos2x+√3/2
=cos60sin2x+cos2xsin60+√3/2
=sin(2x+π/3))+√3/2
T=2π/w=2π/2=π
所以最小正周期是π
（2）sin的原单调增区间是（-π/2+2kπ,π/2+2kπ）减区间（π/2+2kπ,3π/2+2kπ）
f(x)的增区间为（-5π/12+kπ,π/12+kπ）减区间（π/12+kπ,7π/12+kπ）
由图像可知π/2时最小,
代入得最小值0,最大1+√3/2

已知函数f(x)=sinxcosx+(√3)cos²x；(1)求f(x)的最小正周期；(2),求f(x)在区间[-π/6,π/2]上的最大值和最小值。
f(x)=(1/2)sin2x+(√3/2)(1+cos2x)=sin2xcos(π/3)+cos2xsin(π/3)+√3/2=sin(2x+π/3)+√3/2
故最小正周期T=π；
f(x)在区间[-π/6,π...

f(x)=1/2sin2x+√3/2（cos2x+1）
=sin(2x+π/3)+√3/2
1.T=π
2.f(x)max=f(π/3)=1+√3/2
f(x)min=f(-π/6)=√3/2