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已知函数F(x)=2根号3sinXCosX+2Cos^2X%1(X属于R)

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

求采纳,

解1f(x)=cos2x+√3×2sinxcosx =cos2x+√3sin2x =2(1/2cos2x+√3/2sin2x) =2sin(2x+π/6) 故T=2π/2=π 当2kπ-π/2≤2x+π/6≤2kπ+π/2,k属于Z时,y是增函数。 即2kπ-2π/3≤2x≤2kπ+π/3,k属于Z时,y是增函数。 即kπ-π/3≤x≤kπ+π/6,k属于Z时,y是增函数。 故函数...

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

f(x)=√3sinxcosx+cos²x =(√3/2)·2sinxcosx+½(2cos²x-1)+½ =(√3/2)sin2x+½cos2x+½ =sin(2x+π/6) +½ 最小正周期T=2π/2=π

f(x)=sin^2x+√3sinxcosx+2cos^2x =cos^2x+√3/2*2sinxcosx+1 =1/2cos2x+√3/2sin2x+3/2 =sinx(2x+π/6)+3/2 T=2π/2=π 2x+π/6在[2kπ-π/2,2kπ+π/2]上单调递增 x在[kπ-5π/12,kπ+π/6]上单调递增 2)y=sin2x的图像经X轴向左平移π/12个单位得到y=sinx(2x+...

fx=2根号3sinxcosx-cos2x =2(根号3/2sin2x-1/2cos2x) =2sin(2x-派/6) 增区间: 2k派-派/3

题干多题,不能正常作答。

f(x)=2√3sinxcosx+1-2(sinx)^2 =√3sin2x+cos2x =2sin(2x+π/6) 所以最小正周期=2π/2=π -π/2+2kπ

(1).f(x) =√3/2sin2x+1/2cos2x =sin(2x+π/6) T=2π/ω=π (2)将y=sinx的图像向右平移π/6个单位得y=sin(x+π/6),再将y=sin(x+π/6)的图像上的点纵坐标不变,横坐标缩短一倍得y=sin(2x+π/6).

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