已知函数f(x)=1 根号3-2x-x²

f(x)=cos^2x-sin^2x+2(根号3)sinxcosx+1=cos2x+(根号3)sin2x+1=2{(1/2)cos2x+[(根号3)/2]sin2x}+1=2sin(2x+派/6)+1

f(x)=(√3)sinxcosx+cos2x+1f(x)=(√3)(2sinxcosx)/2+cos2x+1f(x)=(√3/2)sin2x+cos2x+1f(x)=(√7/2)[(√3/2)(2/

f(x)=2√3sinxcosx-cos2x=√3sin2x-cos2x=2(sin2x*√3/2-cos2x*1/2)=2sin(2x-π/6)x=π/12;函数f(x)的图象可以由函数y(x)=2

答：f(x)=(1/2)*(cosx)^2+(√3/2)sinxcosx+1=(1/2)*(cos2x+1)/2+(√3/4)sin2x+1=(1/2)[sin2xcosπ/6+cos2xsinπ/6

奇函数证明：f(-x)=log((根号1+X^2)-x)=log(1/X+根号1+X^2)(分子有理化)=-log(X+根号1+X^2)=-f(x)得证

已知函数f(x)=根号3sin2x+cos2x+21求f(x)的最大值及f(x)取得最大值时自变量x集合f(x)=根号3sin2x+cos2x+2=2[(根号3/2)sin2x+(1/2)cos2x]

sin^2X=1/2-(1/2)cos(2X))sof(x)=-1+cos(2x)+sqrt(3)sin2x+1=2sin(2x+pi/6)whenxbelongsto-pi/6,pi/3,2x+pi

f(x)=2√3sinxcosx+2cos^2x-1=√3sin(2x)+cos(2x)=2[sin(2x)(√3/2)+cos(2x)(1/2)]=2[sin(2x)cos(π/6)+cos(2x)

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F(X)=2乘根号3sinxcosx+2cos^2X-1=根号3sin2x+cos2x=2cos(2x-派/3)函数最小正周期=2派/2=派单调递减区间[派/6,2派/3]fx0=6/5,cos(2x

f(x)=1+2(√3)sinxcosx+2(cosx)^2f(x)=2(√3)sinxcosx-[1-2(cosx)^2]+2f(x)=(√3)sin2x-cos2x+2f(x)=2{[(√3)/2

f(x)=2＋√3sin2x＋cos2x=2sin(2x＋π/6)＋21、增区间：2kπ－π/2≤2x＋π/6≤2kπ＋π/2kπ－π/3≤x≤kπ＋π/6则：增区间是：[kπ－π/3,kπ＋π/6]

(1)f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2（√3/2sin2x+1/2cos2x）=2sin（2x+π/6）函数的最小正周期T=π.在区间[0,

f(x)=2根号3sinxcosx+2cos^2x-1=√3sin2x+cos2x=2sin(2x+π/6)kπ-π/3再问：(cos2x)^2=16/25+2√3/5sin2x+3(sin2x)^2

f(x)=1/2cos2x+根号3/2sin2x-1-cos2x=√3/2×sin2x－1/2×cos2x－1＝sin(2x-π/6)－1当sin(2x-π/6)＝1时f(x)有最大值＝0f(a)=-

f(x)=√3sinxcosx+cos2x+1=(√3/2)sin2x+cos2x+1=[(√7)/2][(√3/√7)sin2x+(2/√7)cos2x]+1=[(√7)/2]sin(2x+α)+1

（1）f=[根号3]/2sin2x-cos^2(x)-1/2=√3/2*sin2x-1/2(1+cos2x)-1/2=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1∵x∈[-π/

1、∵1-cos2x=2sin^2∴f(x)=2sinxsinx+2√3sinxcosx+1=1-cos2x+√3sin2x+1=2+2(√3/2*sin2x-1/2*cos2x)=2(cos(π/6

1、f(x)=2sinxsinx+2√3sinxcosx+1=1-cos2x+√3sin2x+1=2sin(2x-π/6)+2对称轴：2x-π/6=kπ+π/2,得x=kπ/2+π/3对称中心：2x-

f(x)=√3sin2x+(1-cos2x)-1=√3sin2x-cos2x=2sin(2x-π/6)最小正周期T=2π/2=π单调增区间：2kπ-π/2=