=(√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+π/6,k∈Z
所以单调递增区间是(kπ-π/3,kπ+π/6),k∈Z
=根号3/2sin2x+cosx平方+1
=根号3/2sin2x+1/2cos2x+3/2
=sin（2x+30°）+3/2
最小正周期t=2π/2=π
单增：2x+30°在{π/2+2kπ，3π/2+2kπ}k∈z
最终答案T=π，区间{π/6+kπ，2π/3+kπ}k∈z
已知函数F(X)=sin²X+√3sinxcosx+2cos²X,X∈R 1 . 求函数的最...
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=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...
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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...
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求助:已知函数f(x)=sin^2x+2√3sinxcosx-1\/2cos2x,x属于R,(1)求f(x...
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解如下图所示
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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...
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