已知函数y等于3减2sin(2X减三分之π)
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y=2sin(2x+π/3)由函数可以看出:振幅:A=2周期:T=(2π)/2=π初相:φ=π/3(x=0时)
1A=2T=2π/2=π初相=π/322x+π/3=π/2+kπ所以对称轴为x=π/12+kπ/2-π/2+2kπ≦2x+π/3≦π/2+2kπ所以-5π/12+kπ≦x≦π/12+kπ所以递增区间为
振幅为2;周期为π;初相为π/3单增区间:kπ-5π/12≦x≦kπ+π/12对称轴:x=﹙1/2﹚kπ+(1/12)π
(0,1)代入原式知sinφ=1/2因为|φ|
最大值 5 ,此时sin(2x-π/3)=1,2x-π/3=π/2+2Kπ,得X=5π/12+Kπ最小值 51,此时sin(2x-π/3)=-1,2x-π/3=-π/2+
f(-x)=f(x)2sin(-wx+θ)=2sin(wx+θ)若-wx+θ=2kπ+wx+θwx=-kπ不成立因此,-wx+θ=2kπ+π-(wx+θ)θ=kπ+π/2所以,可能是:θ=π/2再问:
y=sin^2x+√3sin^2xcosx+2cos^2x你确定那边+√3sin^2xcosx如果是+√3sinxcosx那么y=sin^2x+√3sinxcosx+2cos^2x=1/2(2cos^
sina=3/2sin^2α+sin^2βy=sin^2α+sin^2β=sin^2α+sina-3/2sin^2α=sina-1/2sin^2α-1=
由化简sinx+cosx前分别乘以根号2*sin45.根号2*cos45.,得解根号2sinxy=sinx的平方+根好2*sinx+2令t=sinx-1=
y=sin²x+sinx+cosx+2=(1-cos2x)/2+√2sin(x+л/4)+2=(1/2)*sin(2x+л/2)+√2*sin(x+л/4)+5/2;=(1/2)*sin(2
我列个去,就算我高中毕业到现在已经8年了,我也看的出来1楼的乱说的撒,值域明显是[-2,2]嘛
32/27在原方程上加一个减一个cos^2x化简成cos^3x-cos^2x-cosx+1=0可以令y=cosx.即y^3-y^2-y+1=0然后求导得3y^2-2y-1=0得出y=1或-1/3y=-
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
(1)原式=2[(1/2)sin(x/2)+(根号3/2)cos(x/2)]=2sin[(x/2)+pi/3]所以当[(x/2)+pi/3]=2kpi+pi/2时,y最大值为2解得x=4kpi+pi/
这题目是不是Y=SIn2X2SinX=cosx+3cosX?感觉不对劲啊请完善题目
因为,-π/2
解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数
函数y=2sin(3x+π/6)当函数y取最大值时有3x+π/6=2kπ+π/2即x=2kπ/3+π/9,k∈Z所以x得集合为{x|x=2kπ/3+π/9,k∈Z}
y=sin(1/2x+π/3),x属于R当1/2x+π/3=2kπ+π/2时,y=sin(1/2x+π/3)有最大值1此时x=4kπ+π-2π/3=4kπ+π/3,k∈Z当1/2x+π/3∈【2kπ+