2sinx(-3x π 4)的单调递减

1）f(x)=a·b=-2sin^x+2√3sinxcosx=cos2x-1+√3sin2x=2sin(2x+π/6)-1,它的增区间由（2k-1/2)π

f(x)=2cosx*sin(x+π/3)-根号3sinx^2+sinxcosx=2cosx*(1/2sinx+根号3/2cosx)-根号3sinx^2+sinxcosx=2sinxcosx+根号3c

f(x)=(cosx)^2+sinxcosx=(1/2)sin2x+(1/2)cos2x+1/2=(√2/2)sin(2x+π/4)+1/22kπ-π/2

[-1,-0.5）

f(x)＝sinx＋sin(x－π/3).=sinx+sinxcosπ/3-cosxsinπ/3=3/2*sinx-√3/2*cosx=√3(sinxcosπ/6-cosxsinπ/6)=√3sin(

y'=cosx(1+cos)+sinx(0-sinx)=cosx+cos^2(x)-sin^2(x)=cosx+cos(2x)=2cos^2(x)+cosx-1cosx+2(cosx)^2-1=02(

[π/2,3π/2]再问：f(x)=(x^2-3/2x)e^x的单调增区间再答：这得求导了.f'(x)=(x^2-3/2x)e^x+(2x-3/2)e^x=e^x(x^2-3/2x+2x-3/2)=e

f(x)=sinx-√3cosx=2[(1/2)sinx-(√3/2)cosx]=2[sinxcos(π/3)-cosxsin(π/3)]=2sin(x-π/3)令-π/2+2kπ≤x-π/3≤π/2

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f(x)=2sinx(sinx+cosx)=2sin²x+2sinxcosx=1-cos(2x)+sin(2x)=√2sin(2x-π/4)+1当2kπ-π/2≤2x-π/4≤2kπ

f（x）=sin x-3cos x=2sin（x-π3），因x-π3∈[-43π，-π3]，故x-π3∈[-12π，-π3]，得x∈[-π6，0]，故选D

f(x)=2sin²x+2sinxcosx=1-cos2x+sin2x=1+√2*(√2/2*sin2x-√2/2*cos2x)=1+√2*sin(2x-π/4)令2kπ-π/2≤2x-π/

f(x)=2(sinx-cosx)cosx+1=2sinxcosx-2cos²x+1=sin2x-cos2x=√2sin(2x-π/4)x∈[π/3,3π/4]2x-π/4∈[5π/12,5

f'（x）=1-2cosx＞0x∈（0，π）解得：x∈（π3，π）故答案为：（π3，π）

设x=sinxf（-x）+3f（x）=4*x*√(1-x^2).①设x--sinxf（x）+3f（-x）=4*（-x）*√(1-x^2).②①②分别相加相减得到③④4f（x）+4f（-x）=0.③2f

先化简f(x)=2sinxcosx-cos2x=sin2x-cos2x剩下的,楼主会做了吧如果还不会,请往下看F（X）=√2sin(2x-π/4)所以追小正周期就是π单调增区间算法2Kπ+π∕2

解f(x)=cosx-√3sinx=2[(1/2)cosx-(√3/2)sinx]=2cos(x+π/3)令2kπ≤x+π/3≤2kπ+π得2kπ-π/3≤x≤2kπ+2π/3∵x∈[0,2π],k可

(1)y=sin(π/4-3x)递增2kπ-π/2

1、由公式sinx/cosx=tanx和sinx平方加cosx平方=1列出方程组解sinx=12/13;cosx=5/13.2、当x在区间（0,π）时,令z=x+π/3,z的区间为（π/3,4π/3）

f(x)=2×(1/2sinx-√3/2cosx)=2×(cosπ/3sinx-sinπ/3cosx)=2sin(x-π/3)x∈【-π,0】x-π/3∈【-4π/3,-π/3】x-π/3∈【-π/2