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已知函数F(x)=sinx^2+√3sinxCosx+2Cosx^2,x属于R

f(x)=(1-cos2x)/2+√3sin2x/2+(1+cos2x) =√3sin2x/2+cos2x/2+3/2 =sin(2x+π/6)+3/2 1.最小正周期=2π/2=π 单调增区间2kπ-π/2<2x+π/6<2kπ+π/2,kπ-π/3<x<kπ+π/6, (kπ-π/3,kπ+π/6) 单调增区间2kπ+π/2<2x+π/6<2kπ+3π/2,kπ+π/6<x<kπ+2π/3, (kπ+π/6,kπ+2π/3) 2.y=sin2x(左移π/12)→y=sin(2x+π/6)(上移3/2)→y=sin(2x+π/6)+3/2

答:f(x)=(sinx)^2+√3sinxcosx+2(cosx)^2=1+ √3sinxcosx+(cosx)^2=1+ √3sin(2x)/2+(1+cos2x)/2=3/2+sin(2x)*cos(π/6)+cos(2x)*sin(π/6)=3/2+sin(2x+π/6)故f(x)的最小正周期是π

f(x)=sin^2+√3sinxcosx+2cos^2x=f(x)=√3/2sin2x+cos^2x+1=√3/2sin2x+1/2cos2x+3/2=sin(2x+π/6)+3/21 最小正周期为π.对称轴方程为令sin(2x+π/6)=±1,得x=π/6+kπ,或者x=-π/3+kπ2、当sin(2x+π/6)=1时,f(x0最大 x=π/6+kπ3、y=sinx 所有的横坐标变为原来的一半,得到y=sin2x,向左移动π/12个单位 得到y=sin(2x+π/6),再向上平移3/2个单位

f(x)=sinx+√3sinxcosx+2cosx=(1-cos2x)/2+√3/2(2sinxcosx)+(1+cos2x) =(1-cos2x)/2+√3/2sin2x+(1+cos2x) ∴f(π/12) =(1-cosπ/6)/2+√3/2sinπ/6+(1+cosπ/6) =(1-√3/2)/2+√3/2*1/2+(1+√3/2) =3+√3/2请复核数字计算

解:(1)f(x)=sinx+√3sinxcosx+2cosx=cosx +(√3/2)sin(2x)+1=[1+cos(2x)]/2 +(√3/2)sin(2x) +1=(√3/2)sin(2x)+(1/2)cos(2x) +3/2=sin(2x+π/6) +3/2最小正周期Tmin=2π/2=π2kπ-π/2≤2x+π/6≤2kπ+π/2 (k∈Z)时,函数单调递增,此时kπ-π/3≤x

f(x)=sinx^2+√3sinxcosx+2cosx^2=sinx^2+cosx^2+√3sinxcosx+cosx^2=1+√3sinxcosx+cosx^2=1+√3/2*sin2x+(1+cos2x)/2=1+√3/2*sin2x+1/2+1/2*cos2x=√3/2*sin2x+1/2*cos2x+3/2=sin2xcosπ/6+cos2xsinπ/6+3/2=sin(2x+π/6)+3/2T=2π/2=π2x+π/6∈(2kπ-π/2,2kπ+π/2)2x∈(2kπ-2π/3,2kπ+π/3)x∈(kπ-π/3,kπ+π/6)所以f(x)单调增区间为:x∈(kπ-π/3,kπ+π/6)

f(x)=2sinx^2+sinx*cosx+cosx^2,x∈R=1-cos(2x)+1/2sin(2x)+(1+cos(2x))/2=√2/2sin(2x-π/4)+3/2(1)f(π/12)=(√3+5)/4(2)f(x)的最小值为3/2-√2/2x=-π/8+kπ,k∈Z(3)根据2kπ-π/2≤2x-π/4≤2kπ+π/2,k∈Z得出即可.

(1) f(x)=sinx^2+2sinxcosx+3cosx^2 =sin(2x)+cos(2x)+2 =根号2sin(2x+pai/4)+2所以,f(x)max=根号2+2 x属于R(2)由(1)知, f(x)求导=根号2cos(2x+pai/4)大于0, 得,x在(k/2-3*pi/8,k/2+pi/8)上单调递增

f(x)=sin^2x+√3sinxcosx+cos^2x+cos^2x=1+√3sinxcosx+cos^2x=1+√3/2sin2x+(1+cos2x)/2=√3/2sin2x+1/2cos2x+3/2=sin(2x+π/6)+3/2T=2π/2=π令2kπ-π/2<=2x+π/6<=2kπ+π/2解得,kπ-π/3<x<kπ+π/6单调增区间为[kπ-π/3<x<kπ+π/6]sin2x向左移π/12个单位,得到sin2(x+π/12)=sin(2x+π/6)再向上移动3/2单位,得到sin(2x+π/6)+3/2谢谢采纳!

(1)f(x)=(cosx)^2-(sinx)^2+2√3sinxcosx =cos2x+√3sin2x =2(√3/2sin2x+1/2cos2x) =2sin(2x+π/6)所以f(π/12)=2sin(π/12*2+π/6)=2sinπ/3=√3(2)最小正周期T=2π/2=π最大值=2

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