函数f(x)等于3sin(2x-y)满足f(x π.6)

f(x)=(1/2)sin2x+(√3/2)cos2x+(1/2)cos2x+(√3/2)sin2x=[1+(√3)]/2(sin2x+cos2x)=[1+(√3)]/√2(1/√2*sin2x+1/

1、由于函数g(x)=sin(2(x－a)+π/3)为偶函数,所以g(x)的图像关于y轴对称,即函数g(x)当x=0时取得最值,所以g(0)=±1,解得sin(π/3－2a)=±1,sin(2a－π/

令x=π/3代入f（（π/3）+x)=f（（π/3）-x)f(0)=sinA+1=f(2π/3)=sin((4π/3)-A）+1sinA=sin((4π/3)-A）=sin(4π/3)cosA-cos

(1)f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x=√3-1/2sin2x-√3/2cos2x=√3-sin(2x+π/3)∴最小正周期T=2π/2=π单调增区间：π/2+2

y＝sin(x+π3)sin(x+π2)＝(sinxcosπ3+cosxsinπ3)cosx=12sinxcosx+32cos2x＝14sin2x+32•1+cos2x2=34+12sin(2x+π3

∵f(x)＝2sin2x−23sinxsin(x−π2)=2sin2x+23sinxcosx=1−cos2x+3sin2x=1+2sin(2x−π6)∵0＜x＜2π3∴−π6＜2x−π6＜7π6∴−1

f(x)=sin²(x)+(√3)sin(x)cos(x)+2cos²(x)=3/2+√3/2sin2x+1/2cos2x=3/2+sin(2x+π/6)函数f(x)的最小正周期T

由f(x)＝3sinx+cosx＝2sin(x+π6)⇒f(x)max＝2．故答案为：2

π/3+θ=kπ,θ=kπ-π/3,k是整数

f(x)=2sin(-3x+π/3+θ)是奇函数则f（0）=0故π/3+θ=kπ得到θ=kπ-π/3（k是整数）

y=sinx^2+根3sinxcosx+2cosx^2=-1/2(1-2sinx^2)+1/2根3*2sinxcosx+2cosx^2-1+3/2=-1/2cos2x+二分之根3倍sin2x+cos2

f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T

∵x∈[0，π3]，∴π3≤x+π3≤2π3，根据正弦函数的性质得，32≤sin(x+π3)≤1，则3≤2sin(x+π3)≤2，∴f（x）的值域是[3，2]．故答案为：[3，2]．

fx=2cosx(0.5sinx+根号3/2cosx)-根号3sin*2x+sinxcosx=2sinxcosx+根号3（cos*2x-sin*2x）=sin2x+根号3cos2x=2sin（2x+派

(1)F(X)=√3sin2x+2sin²x=√3sin2x+1-cos2x=2(√3/2sin2x-1/2cos2x)+1=2sin(2x-π/6)+1F(x)的最小正周期T=2π/2=π

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

f（x）=sin^2x+2√3sinxcosx+3cos^2x=1+√3sin2x+2cos^2x-1+1=√3sin2x+cos2x+2=2(sin2x*√3/2+cos2x*1/2)+2=2sin

如果我知识点还没忘记的话,应该是（n+1/12）π,其中n=0,1,2,3,4,.因为sinx根据（2nπ+π/2）对称,列出等式2x+π/3=2nπ+π/2即可

答案：9再问：图中不是只有4个零点？再答：还有两个端点！~再问：看懂了！两个图像有两个零点很接进！再答：祝你开心~