f x sinx 根号3cosx

(3/2)cosx-(√3/2)sinx=√3[(√3/2)cosx-(1/2)sinx]=√3[sin(π/3)cosx-cos(π/3)sinx]=√3sin(π/3-x)=-√3sin(x-π/

/>tanx=√3则x=kπ+π/3,k∈Z（1）k是偶数,则sinx=√3/2,cosx=1/2（2）k是奇数则sinx=√3/2,cosx=1/2(sinx+cosx)/(sinx-cosx)的值

用sinx2+cosx2=1再问：可以帮我把顺序写出吗。再问：谢谢。再答：就是3cos2x+cos2x=1再答：然后cosx=正负0.5就行了再问：就只有后面这两个步骤吧再问：还是前面那个也要再问：s

sinx+根号3cosx=22(sinx/2+根号3cosx/2)=2sinx/2+根号3cosx/2=1sinxcosπ/3+cosxsinπ/3=1sin(x+π/3)=1所以x+π/3=2kπ+

（1）3根号15sinx+3根号5cosx=6根号5((根号3)/2sinx+1/2cosx)=6根号5(cos10°sinx+sin30°cosx)=(6根号5)sin(x+30°)（2）根号3si

sinx+cosx=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)sinx-cosx=√2(sinxcosπ/4-cosxsinπ/4)=√2sin(x-π/4)sinx

=2*(根号3sinx/2+cosx/2)=2(sinx*cos30°+cosxsin30°）=2sin（x+30°)

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

=(sinx)^2+√3sinxcosx=(1-cos2x)/2+√3/2(2sinxcosx)=1/2-1/2cosx+√3/2sin2x=1/2+sin(-π/6)cos2x+cos(-π/6)s

y=sinxcosx-√3(cosx)^2+√3/2=1/2*sin2x-√3*[(1+cos2x)/2]+√3/2=(sin2x)/2-(√3/2)*cos2x-√3/2+√3/2=√[(1/2)^

（1）解析：函数f(x)=√3cos²x+sinxcosx=√3(cos2x+1)/2+1/2sin2x=sin(2x+π/6)+√3则函数的最小正周期为π,图像的对称轴方程x=kπ+π/6

π/6+2kπ（k属于Z）

y=√3cosx+sinx=2[(√3/2)cosx+(1/2)sinx]=2(sinπ/3cosx+cosπ/3sinx)=2sin(x+π/3)所以它的最小正周期=2π；图像的对称轴为:x=kπ+

不用画图值域就是[-2,2]因为y=cosx的值域是[-1,1]

y=cosx(sinx-√3cosx)+√3/2=cosxsinx-√3(cosx)^2+√3/2=sin2x/2-√3cos2x/2=sin(2x-π/3)当x=-π/2时y=sin(-π-π/3)

Y=sinx*cosx*√3+cos^2x=√3/2*sin2x+1/2*cos2x+1/2=sin(2x+∏/6)+1/2.

不管怎么写,都是有意义的.它只是函数,不存在成立不成立的问题.

------a^2+c^2-b^2a^2+c^2-b^2a^2+c^2-accosB=-------------=-------------=-------------2ac2ac2ac2ac-ac1

没错,f(x)=2sin(2x+π/6)周期T=2π/2=π因为-1≤sin(2x+π/6)≤1f(x)max=2f(x)min=-2

cosx=根号3/2时x=2kπ+π/6或x=2kπ-π/6,k为整数所以不等式cosx>根号3/2解为：2kπ-π/6