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已知函数F(x)=根号3sin²x+sinxCosx,x属于【π/2,π】

f(x)=√3sinx+sinxcosx=√3/2(1-cos2x)+1/2sin2x=√3/2+sin(2x-π/6)f(x)=0,sin(2x-π/6)=-√3/2 ,2x-π/6=π/3 ,x=π/4 ,2x-π/6=2π/3 ,x=5π/12

1)f(x)=sinx(√3 sinx+cosx)=0有sinx=0, 得x=π及√3 sinx+cosx=0, 得tanx=-√3/3, 即x=5π/62) )f(x)=√3 (1-cos2x)/2+1/2*sin2x=√3/2+(1/2*sin2x-√3/2 cos2x)=√3/2+sin(2x-π/3)因为2π/3=<(2x-π/3)<= 5π/3所以有当2x-π/3=2π/3时,fmax=f(π/2)=√3当2x-π/3=3π/2时, fmin=f(11π/12)=√3/2-1

f(x)=根号3sinx+sinxcosx-根号3/2=根号3(1/2-cos(2x)/2)+1/2*sin2x-根号3/2=1/2*(sin2x-根号(3)cos2x)=根号(1+3)/2*sin(2x-π/3)=sin(2x-π/3)f(A)=f(B)=1/2sin(2A-π/3)=sin(2B-π/3)=1/2,A

f(x)=sinx(√3sinx-cosx) =sinx[2sin(x-π/6)]sinx[2sin(x-π/6)]=0因为,π/3≤(x-π/6)≤5π/6,所以,sin(x-π/6)≠0sinx=0==>x=π

(1) f(x)=(1-cos2x)/2+√3/2*sin2x+1+cos2x=√3/2*sin2x+1/2*cos2x+3/2=sin(2x+π/6)+3/2所以f(x)的最小正周期为π在(kπ-π/3,kπ+π/6)上单调递增,在(kπ+π/6,kπ+2π/3)上单调递减(2) f(x)可以由函数y=sin2x的图像向左平移π/12,向上平移3/2得到

f(x)=-√3(sinx)^2+sinxcosx =-√3/2*(1-cos2x)+1/2*sin2x =1/2*sin2x+√3/2*cos2x-√3/2 =sin(2x+π/3)-√3/21最小正周期为 π2因为 0≤x≤π/2,所以 π/3≤2x+π/3≤π+π/3所以 -√3/2≤sin(2x+π/3)≤1因此,函数的值域是 【-√3,1-√3/2】.

sqr(3)是根号3的意思f(x)=[(sqr(3)/2)*(1-cos2x)]+(1/2)*sin2x=(1/2)*sin2x-(sqr(3)/2)*cos2x+sqr(3)/2=sin(2x-π/3)+sqr(3)/2 2π/3≤2x-π/3≤5π/3 所以函数的零点是2x-π/3=π,x=2π/3, 当2x-π/3=2π/3,即x=π/2时,取最大值 0当2x-π/3=3π/2,即x=11π/6时,取最小值 -1+sqr(3)/2

f(x)=√3sin^2 x+sinxcosx-(√3/2)(x∈R)=√3*[(1-cos2x)/2]+(1/2)sin2x-(√3/2)=(√3/2)-(√3/2)cos2x+(1/2)sin2x-(√3/2)=(1/2)sin2x-(√3/2)cos2x=sin[2x-(π/3)](1)f(π/4)=sin[(π/2)-(π/3)]=sin(π/6)=1/2(2)当x∈(0,π)时,2x∈(0,2π)2x-(π/3)∈(-π/3,5π/3)所以,f(

f(x) = √3sinx + sinx cosx=√3 (1 - cos2x)/2 +1/2sin2x=√3/2-√3/2cos2x +1/2sin2x=√3/2- sinπ/3 cos2x + cosπ/3 sin2x= √3/2 + sin(2x-π/3) f(2Pai/3)=根号3/2+sinPai/3=根号3最大值=根号3/2+1

f(x)=2根号下3sin(π-x)sin(π/2-x)-2cos(π+x)cosx+2 =2cosx(根号下3sinx+cosx)+2 =根号下3sin2x+cos2x+3 =2sin(2x+π/6)+3最小正周期=2π÷2=π 单调递减区间:2kπ+π/2 追问: 第一小题的第二行是怎么推出第三行的?第三行的“根号下”是3还是3sin2x? 追答: 开括号=2根号下3sinxcosx+2cosx-1+3 =根号下3sin2x+cos2x+3是根号下3 评论0 0 0

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