f x=3 sin平方x

f(x)=√3sin²x+sinxcosx=√3[(1-cos2x)/2]+1/2sin2x=1/2sin2x-√3/2cos2x+√3/2=sin(2x-π/3)+√3/2∵x∈[π/2,

(1)f(x)=2sin(2x+π/3)+2由2x+π/3=kπ+π/2,k∈Z得2x=kπ+π/6,k∈Z对称轴方程为x=kπ/2+π/12,k∈Z(2)g（x）=f（x）+m=2sin(2x+π/

f(x)=(1/2)[(cosx)^2-(sinx)^2]-(1/2)sinxcosx=(1/2)cos2x-(1/2)sin2x=1/2[cos2x-sin2x]=-√2/2sin(2x-π/4)周

周期等于2派.g(x)=2sinx;基函数再问：有过程吗？？再答：这可以看出来，还要过程吗，，，，周期等于2派/x前的数1===2派；；g(x)=2sint(x+pi/3+p1/3)=2sinx;si

1、最小正周期T=2π/2=π;最大值=2×1+2=4；2、单调递增式时-π/2+2kπ≤2x+π/3≤π/2+2kπ(k∈Z)-5π/6+2kπ≤2x≤π/6+2kπ(k∈Z)-5π/12+kπ≤x

fx=4cos²x-2+1-cos²x-4cosx=3cos²x-4cosx-1令t=cosx则-1≤t≤1即求[3t²-4t-1]的最值

答：f(x)是R上奇函数：f(-x)=-f(x),f(0)=0x>0,f(x)=x^2-3x+2x<0,-x>0：f(-x)=(-x)^2-3(-x)+2=x^2+3x+2=-f(x

fx=2cosxsin（x+π/3）-√3sin^2x+sinxcosx+1=2cosx(√3/2cosx+1/2sinx)-√3sin^2x+sinxcosx+1=√3cos^2x-√3sin^2x

答：f(x)=2sin(x-π/3)cosx+sinxcosx+√3(sinx)^2=sin(x-π/3+x)+sin(x-π/3-x)+sinxcosx+(√3/2)(1-cos2x)=sin(2x

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]=cos(2x-π/3)+2sin(x-π/

f(x)=[1-cos(2x)]/2+sin(2x)+3[1+cos(2x)]/2=sin(2x)+cos(2x)+2=√2sin(2x+π/4)+2.周期T=kπ,k∈Z且k≠0.最小正周期为π.

f(x)=(√3/2)sin2x-(1/2)[(cosx)^2-(sinx)^2]-1=(√3/2)sin2x-(1/2)cos2x-1=sin(2x-π/6)-1f(x)的最大值是0,最小值是-2,

(1)f（x）=（sinx）^2+2√3sinxcosx+3（cosx）^2=√3sin2x+cos2x+2=2sin(2x+π/6)+2,f(x)的增区间由（2k-1/2)π

等一下再答：

f(x)=2sin(x-π/6)cosx+2cos²x=(2sinxcosπ/6-2cosxsinπ/6)cosx+2cos²x=√3sinxcosx-cos²x+2co

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=(1/2)cos2x+(√3/2)sin2x+(cos(π/2)-cos2x)=-(1/2)cos2x+(√3/2)sin

T=2π/2=π[-1,1]最大值为1,最小值为-1

f(x)=cos²2x-sin²2x+sin4x=cos4x+sin4x=√2[(√2/2)cos4x+(√2/2)sin4x]=√2sin(4x+π/4)所以,最小正周期T=2π

解答；f(x)=sin(2x+3分之π)∴sin(2x+π/3)=-3/5∵x∈(0,π/2)∴2x+π/3∈(π/3,4π/3)∵sin(2x+π/3)

解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数