=1/2(1-cos2x)+√3/2*sin2x+1+cos2x
=√3/2sin2x+1/2cos2x+3/2
=sin(2x+π/6)+3/2
函数的最小正周期T=2π/2=π
由2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈Z
得2kπ-2π/3≤2x≤2kπ+π/3,k∈Z
∴kπ-π/3≤x≤kπ+π/6,k∈Z
∴函数递增区间为
[kπ-π/3,kπ+π/6],k∈Z
=（1-cos2X）/2 + √3/2 sin2X + 1+cos2X
=√3/2 sin2X + 1/2cos2X + 3/2
=sin(2X+π/6) +3/2
最小正周期T=2π/2=π
单调递增区间 (k∈Z)
2X+π/6∈[2kπ-π/2，2kπ+π/2]
2X∈[2kπ-π2/3，2kπ+π/3]
X∈[kπ-π/3，kπ+π/6]
已知函数F(X)=sin²X+√3sinxcosx+2cos²X,X∈R 1 . 求函数的最...
F(x)=sin²x+√3sinxcosx+2cos²x =(√3\/2)sin2x+1+(1+cos2x)\/2 =(√3\/2)sin2x+(1\/2)cos2x+3\/2 =sin(2x+π\/6)+3\/2 所以最小正周期是T=2π\/2=π 令2kπ-π\/2＜2x+π\/6＜2kπ+π\/2,k∈Z 2kπ-2π\/3＜2x＜2kπ+π\/3,k∈Z kπ-π\/3＜x＜kπ...
已知函数f(x)=sin²x+根号3sinxcosx+2cos²x,x∈R
=1\/2(2cos²x-1)+1\/2+(sin²x+cos²x)+√3\/2(2sinxcosx)=1\/2cos2x+√3\/2sin2x+3\/2 =sin(2x+π\/6)+3\/2 所以可得：最小正周期T=2π\/2=π 单调增区间为：[kπ-π\/3,kπ+π\/6] (注：k为任意整数)x∈【0，π\/2】时求函数的值域 当x=π\/6时有...
已知函数f(x)=sin²x+√3xcosx+2cos²x,x属于R
（1）f(x)=sin²x+√3sinxcosx+2cos²x=√3sinxcosx+2cos²x+sin²x =(√3sin2x)\/2+(1+cos2x)\/2+1 =sin(2x+π\/6)+3\/2,最小正周期π， 由2kπ-π\/2≤2x+π\/6≤2kπ+π\/2, k∈Z,得，kπ-π\/3≤x≤kπ+π\/6,由2kπ+π\/2≤2x+π\/6≤...
已知函数f(x)=sin²x+根号3sinxcosx+2cos2x
f（x）=sin²x+根号3sinxcosx+2cos²x =1+（1\/2）（√3sin2x+1+cos2x)=3\/2+sin(2x+π\/6),(1)最小正周期为π，增区间由（2k-1\/2)π<2x+π\/6<(2k+1\/2)π,k∈Z确定，（2k-2\/3)π<2x<(2k+1\/3)π,即(k-1\/3)π<x<(k+1\/6)π.（2）f(x)的图像可由...
已知函数f(x)=sin²x+√3sinxcosx,求函数f(x)的最小正周期和单调增...
f(x)=sin²x+√3sinxcosx=0.5 0.5cos2x+0.5√3sin2x=0.5+sin(2x-π\/6)最小正周期2π\/2=π，单调增区间2kπ- π\/2<2x-π\/6<2kπ+π\/2,kπ- π\/6<x<kπ+2π\/3
求助:已知函数f(x)=sin^2x+2√3sinxcosx-1\/2cos2x,x属于R,(1)求f(x...
求助：已知函数f(x)=sin²x+2(√3)sinxcosx-(1\/2)cos2x,x∈R,(1)求f(x)的最小正周期和值域；(2)若x。(0≦x。≦π\/2）为f(x)的一个零点，求sin2x。的值。解：(1)。f(x)=(1-cos2x)\/2+(√3)sin2x-(1\/2)cos2x=(√3)sin2x-cos2x+1\/2 =2[(√3\/2)sin2x-(1...
已知函数f(x)=sin²x+√3sinxcosx (1)求f(x)的最小正周期(2)单调递增...
解如下图所示
f(x)=sin²x+根号3sinxcosx+2cos²x
f(x)=sin²x+√3sinxcosx+2cos²x =1+cos²x+(√3\/2)sin2x =1+(1+cos2x)\/2+(√3\/2)sin2x =(√3\/2)sin2x+(1\/2)cos2x+3\/2 =sin(2x+π\/6)+3\/2 最小正周期 T=2π\/2=π -π\/2+2kπ≤2x+π\/6≤π\/2+2kπ -π\/3+kπ≤x≤π\/6+kπ 单调递增...
...=sin²x+√3sinxcosx+½1)求函数最小正周期2)求函
f(x)=sin²x+√3sinxcosx+½=-（1\/2-sin²x）+√3sinxcosx+1 =-1\/2cos2x+√3\/2sin2x+1 =sin(2x- ∏\/6)+1 1)T=2∏\/2=∏ 2)由三角函数的值域可得 f(x)的值域[0,2]当f(x)=0时,即sin(2x- ∏\/6)=-1 2x- ∏\/6=- ∏\/2+2k∏ 解得 x=-∏\/6+k...
已知函数f(x)=cos²x+根号3sinxcosx+1,x∈R
f(x)=cos²x+√3sinxcosx+1 =(cos2x+1)\/2+√3\/2*sin2x+1 =(√3\/2*sin2x+1\/2*cos2x)+3\/2 =sin(2x+π\/6)+3\/2 1，最小正周期T=2π\/2=π 最大值f(x)max=1+3\/2=5\/2 最小值f(x)min=-1+3\/2=1\/2 2，令2kπ-π\/2≤2x+π\/6≤2kπ+π\/2，得：kπ-π...
