求函数y=√3sin2x-2sin²x 1的最小正周期-搜答案-百度作业网

y=sin²x+√3/2sin2x+2cos²x=√3/2sin2x+cos²x+1=√3/2sin2x+1/2(2cos²x-1)+3/2=√3/2sin2x

y=e^sin2x复合函数求导：y′=e^sin2x*cos2x*2=2cos2x*e^sin2x

求函数导数 y=sin2x

y'=2cos2x

可以根据：asinα+bcosα=√(a^2+b^2)*sin(α+θ),其中θ由a,b的符号和tanθ=b/a确定具体到这题,就是：y=√[(√3+2)/2]^2+(1/2)^2*sin(2x+θ)

y=(x²+3x)^(sin2x)lny=sin2x·ln(x²+3x)y'/y=ln(x²+3x)·2cos2x+sin2x·(2x+3)/(x²+3x)y'

sin2x＞0;∴0+2kπ＜2x＜π/2+2kπ(k∈Z)0+kπ

y=2(sinx)^2+2cosx-3=2-2(cosx)^2+2cosx-3=-2(cosx)^2+2cosx-1=-2(cosx-1/2)^2-1/2当cosx=1/2时,函数有最大值1/2,此时

y=sin2x-sin(2x-π/3)=2cos(π/3)sin2x-[sin2xcos(π/3)-cos2xsin(π/3)]=sin2xcos(π/3)+cos2xsin(π/3)=sin(2x+

令t=sinx+cosx=√sin（2x+π/4）所以t属于[-√2,√2]sin2x=2sinxcosx=(sinx+cosx)^2-1所以y=t+（t*t-1）/2+2=0.5t*t+t+1.5t

化简得：y=sin（2x+π/3）+根号3/2周期是π,增区间是（kπ-5π/12,kπ+π/12)(k属于z)

y=√3cos²x+1/2sin2x=√3/2(cos2x+1)+1/2sin2x=√3/2cos2x+1/2sin2x+√3/2=sin(π/3)cos2x+cos(π/3)sin2x+√

由y=sin2x+sin2x+3cos2x=1+sin2x+2cos2x=1+sin2x+（1+cos2x）=2sin(2x+π4)+2（1）当sin(2x+π4)＝−1时，y最小=2-2，此时，由2

（1）∵y=sin2x+sin2x+3cos2x=sin2x+cos2x+2=2sin（2x+π4）+2，∴当2x+π4=2kπ-π2（k∈Z），即x=kπ-3π8（k∈Z）时，f（x）取得最小值2-

2sin2x+√3>0,且9-x^2>=0.得sin2x>-√3/2,且|x|

sin2x+1/2>0sin2x>-1/2=sin(2kπ-π/6)=sin(2kπ+7π/6)所以2kπ-π/6

sinx+cosx=√2(2/√2sinx+2/√2cos)=√2sin(x+45),令sinx+cosx=y,则y的取值[-√2,√2](sinx+cosx)^2=1+2sinxcosxsin2x=

y=[sin^2(x)+cos^2(x)]+sin(2x)+2cos^2(x)=1+sin2x+(1+cos2x)==sin2x+cos2x+2=根号(2)*sin(2x+π/4)+2当2x+π/4=

y=2sin(2x-π/3)最大值为2最小值为-2T=2π/2=π