y=3sin(2x π 4),并求出是Y最大值和最小值的X的集合.
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函数的周期T=2πω=2π2=π,由-π2+2kπ≤2x+π3≤π2+2kπ,解得−5π12+kπ≤x≤π12+kπ,即函数的递增区间为[−5π12+kπ,π12+kπ],k∈Z,由2x+π3=π2+
∵y=sin(2x+π3),∴由2kπ−π2≤2x+π3≤2kπ+π2,k∈Z.得kπ-5π12≤x≤kπ+π12,k∈Z.∴当k=0时,递增区间为[0,π12],当k=1时,递增区间为[7π12,π
最大值-1,最小值-3x/3-π/4=π/2+2kπ时,即x=3π/4+2kπ时sin最大=1y最大1-2=-1x/3-π/4=-π/2+2kπ时,即x=-π/4+2kπ时sin最小=-1y最小-1-
振幅:3;周期:2π除以|x的系数|=4π;初相:-π/4;减区间:2kπ+π/2≤x/2-π/4≤2kπ+3π/2,解得:[4kπ+3π/2,4kπ+7π/2],其中k是整数;最小值是-3,此时x/
当1/2x+π/3=2kπ+π/2时,取得最大值,即y=1/2,x∈{x|x=4kπ+π/3,k∈z};当1/2x+π/3=2kπ-π/2时,取得最大值,即y=-1/2,x∈{x|x=4kπ-5π/3
y=3sin(2x+4分之派)最小正周期为2π/2=π单调减区间:2x+π/4∈[2kπ+π/2,2kπ+3π/2]x∈[kπ+π/8,kπ+5π/8]所以单调减区间为;[kπ+π/8,kπ+5π/8
(1)最大值ymax=-1,x/3-π/4=π/2+2kπx/3=3π/4+2kπx=9π/4+6kπ所以,最大值时,x∈{x|x=9π/4+6kπ,k∈Z}(2)最小值ymin=-3x/3-π/4=
y=2sin(2x+π/3)-π/6≤x≤π/6-π/3≤2x≤π/30≤2x+π/3≤2π/3有:0≤sin(2x+π/3)≤1所以:0≤2sin(2x+π/3)≤2即:0≤y≤2可见:y的最大值是
y=﹙sinx)的平方+[sin(x+π/3)]的平方+[sin(x-π/3)]的平方sin(x﹢π/3)=sinxcosπ/3﹢cosxsinπ/3=sinx/2﹢√3cosx/2sin(x﹣π/3
y=sinx的图像向左平移π/4个单位,得到y=sin(x+π/4)将所有点的纵坐标不变,横坐标变为原来的1/2,得到y=sin(2x+π/4)将所有点的横坐标不变,纵坐标变为原来的1/2,得到y=1
y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos
y∈[1,3]当y=1时,sin(x+π/3)=-1,x+π/3=2kπ-π/2,x=kπ-5π/12,k∈Z当y=3时,sin(x+π/3)=1,x+π/3=2kπ+π/2,x=kπ+π/12,k∈
再问:π/2+2kπ得5π/12+kπ是怎样的出来的再答:sin(2x-π/3)的括号,等于π/2+2kπ时,则正弦值等于+1.就是:2x-(π/3)=π/2+2kπ。2x=(5π/6)+2kπ,x=
sinx的减区间是(2kπ+π/2,2kπ+3π/2)所以这里2kπ+π/2
令t=2x-π/12,则2x+π/6=2t+π/4,所以y=2sin(2x+π/6)+2sin(2x-π/12)=2sin(t+π/4)+2sint=√2sint+√2cost+2sint=(√2+2
配方y=-2(sinx-1/2)^2+3/2当x=-π/6时取的最大值1当x=π/2时取的最小值-1/2
如果你看着sinx不习惯,可以用换元法计算首先令sinx=t,即y=-2t^2+2t+1y=-2(t-1/2)^2+3/2因为x∈{-π/6,3π/4},所以-1/2≤sinx≤1,即-1/2≤t≤1
1、y=(cos^2x+sin^2x)^2-2cos^2xsin^2x=1-1/2(sin2x)^2=1-1/4(1-cos4x)=3/4+1/4cos4x周期T=2pi/4=pi/22、y=(根3/
sin^x,3cos^x是平方吗?如果是的话可以这样解y=sin^2(x)+2sinxcosx+3cos^2(x)=sin^2(x)+cos^2(x)+sin(2x)+2cos^2(x)=sin(2x
当2x-π/6=kπ+π/2时,函数取最值,此时x=kπ/2+2π/3,k为整数因为函数振幅为1,所以最大值为1,最小值为-1取最大值时x=kπ+π/3最小值时x=kπ-π/6