3sinx cosx=2sin(x 6)

f(x)=2cosxsin(x+60°)-√3sin²x+sinxcosx=2cosx(sinxcos60º+cosxsin60º)-√3sin²x+sinxc

f(x)=cos^2x-sin^2x+2(根号3)sinxcosx+1=cos2x+(根号3)sin2x+1=2{(1/2)cos2x+[(根号3)/2]sin2x}+1=2sin(2x+派/6)+1

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

f(x)=√3(sin^2x-cos^2x)-2sinxcosx=-√3cos2x-sin2x=-2sin(2x+π/3)1.求最小正周期T=π2.设x∈[-π/3,π/3],求函数的值域和单调区间-

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

f(x)=sin^2x+2sinxcosx+3cos^2x=1+sin2x+2cos²x=2+sin2x+cos2x=2+√2sin（2x+π/4）所以f（x）周期为Kπ（2）先纵向拉长√2

f(x)=-√3sin2x+1/2sin2x+√3/2=(-√3+1/2)sin2x+√3/2因为sin2x的最小正周期为2π/2=π因为-√3+1/2小于0所以单调递增区间为（π/4+kπ,3π/4

sin^2x+2sinxcosx-3cos^2x=12sinxcosx-4cos^2x=0sinxcosx=2cos^2xcosx（sinx-2cosx）=0（1）cosx=0,sinx=±1（2）s

除以cos^2x得原式=tan^2x+2tanx-3=-15/4

y=2cosxsin(x+π/3)-根号3*(sin^2)x+sinxcosx,后两项先提出一个sinx,然后括号内部分用叠加原理,得到y=2cosxsin(x+π/3)+2sinxcos(x+π/3

sin^2X=1/2-(1/2)cos(2X))sof(x)=-1+cos(2x)+sqrt(3)sin2x+1=2sin(2x+pi/6)whenxbelongsto-pi/6,pi/3,2x+pi

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

sinx+2cosx=0sinx=-2cosxsin²x+cos²x=15cos²x=1cos²x=1/52sin²x-3sinxcosx=8cos&

1、∵1-cos2x=2sin^2∴f(x)=2sinxsinx+2√3sinxcosx+1=1-cos2x+√3sin2x+1=2+2(√3/2*sin2x-1/2*cos2x)=2(cos(π/6

1、f(x)=2sinxsinx+2√3sinxcosx+1=1-cos2x+√3sin2x+1=2sin(2x-π/6)+2对称轴：2x-π/6=kπ+π/2,得x=kπ/2+π/3对称中心：2x-

移项sin^2x+2sinxcosx-3cos^2x+2=0sin^2x+2sinxcosx-3cos^2x+2(sin^2x+cos^2x)=03sin^2x+2sinxcosx-cos^2x=0(

万能公式知道吧是要记的哦...sin2x=2tanx/(1+tan^2x)=4/5cos2x=(1-tan^2x)/(1+tan^2x)=-3/5原式=2sin^2x-1+1+sin2x/2-3(2c

最大值2,最小值－2.通过化简,化成一个三角函数式,即可知道结果.请参考图片.

f(x)=√3sin2x+(1-cos2x)-1=√3sin2x-cos2x=2sin(2x-π/6)最小正周期T=2π/2=π单调增区间：2kπ-π/2=