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已知函数F(x)=根号3sinx%Cosx,x属于R,若F(x)大于...

f(x)=√3sinx - cosx, x∈R, =2sin(x - π/6 + 2kπ ) ∵ f(x)≥1, ∴ 2sin(x - π/6 + 2nπ ) ≥1 ∴ sin(x - π/6 + 2nπ ) ≥1/2 ∴ π/6+2mπ ≤ x - π/6 + 2nπ ≤ 5π/6+2mπ (这个,你可以从三角函数的图像看出来的) ∴ π/3+2kπ ≤ x ≤ π+2kπ 其中,k,m,n=1,...

函数f(x)=3sinx-cosx=2sin(x-π6),因为f(x)≥1,所以2sin(x-π6)≥1,所以,2kπ+π6≤x?π6≤2kπ+5π6 k∈Z所以f(x)≥1,则x的取值范围为:{x|2kπ+π3≤x≤2kπ+π,k∈Z}故选:B

f(x)=根号下3sinx×cosx–cos²x–½ =(根3)/2*sin2x-1/2*(cos2x+1)-1/2 =(根3)/2*sin2x-1/2*cos2x-1 =sin(2x-pi/6)-1

f(x)=根号3sinx+cosx =2[sinxcos(π/6)+cosxsin(π/6)] =2sin(x+(π/6)) 而x∈[π/3,2π/3], 所以:π/2

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+...

解: f(x)=√3sinx+cosx+1 =2[(√3/2)sinx+(1/2)cosx]+1 =2sin(x+π/6)+1 最小正周期T=2π/1=2π sin(x+π/6)=1时,f(x)有最大值f(x)max=2+1=3 sin(x+π/6)=-1时,f(x)有最小值f(x)min=-2+1=-1 函数的值域为[-1,3]

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