已知函数y=2sin(2x π 3)求单调减区-搜答案-百度作业网

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)

最大值y=2*1+3=5最小值y=2*（-1）+3=1令2kπ-π/2≤x-π/3≤2kπ+π/2得2kπ-π/6≤x≤2kπ+5π/6,k为整数令2kπ+π/2≤x-π/3≤2kπ+3π/2得2kπ

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+

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

y=2sin(π/6-1/3x)=-2sin（1/3x-π/6）得到T=2π/（1/3）=6π令1/3x-π/6=kπ+π/2得到x=3kπ+2π所以函数的对称轴是x=3kπ+2π令1/3x-π/6=

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

y=3sin(π/6-2x)=--3sin（2x-π/6）(与y=3sin（2x-π/6）的单调区间相反)令-π/2+2kπ

模型y=Asin(ωX+ψ)振幅A=4周期T=2π/ω=4π最大值=A=4最小值=-A=-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

因为,-π/2

（1）令3x+6/π=π/2+2kπ,k取整数,》》》（2）3x+6/π属于（π/2+2kπ,3π/2+2kπ）,k取整数》》》》3x+6/π=2kπ,k取整数》》》（3）x不变y缩小1/2并上移1个

y=sin(2x-π/3)+根号3cos2x=sin2xcosπ/3-cos2xsinπ/3+2sinπ/3cos2x=sin2xcosπ/3+cos2xsinπ/3=sin(2x+π/3)1.振幅=

x∈[-2π/9,π/6]3x+π/3∈[-π/3,5π/6]sin(3x+π/3)∈[-√3/2,1]2sin(3x+π/3)∈[-√3,2]函数的最大值=2函数的最小值=-√3

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}

1、定义域是Rx系数是1所以T=2π/1=2π2、五点法即sin里取0,π/2,π,3π/2,π则x-π/3=0,x=π/3,sin(x-π/3)=0x-π/3=π/2,x=5π/6,sin(x-π/