已知Y=SIN(2X加六分之π X大于等于负六分之π小于等于三分之π求最值
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单调增区间是:2kPai-Pai/2
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
y=sinx的减区间应该是[π/2+2kπ,3π/2+2kπ](/是分数线的意思)所以y=2sin(2x+π/6)的减区间是2x+π/6∈[π/2+2kπ,3π/2+2kπ]即π/2+2kπ=
y=sinx的减区间应该是[π/2+2kπ,3π/2+2kπ](/是分数线的意思)所以y=2sin(2x+π/6)的减区间是2x+π/6∈[π/2+2kπ,3π/2+2kπ]即π/2+2kπ=
x∈【六分之π,三分之二π】2x∈【π/3,4π/3】2x-6分之π∈【π/6,7π/6】2x-6分之π=7π/6时,y有最小值=-1/22x-6分之π=π/2时,y有最大值=1所以:y的值域是【-1
x=-32y=54
后面的sin应该有平方(1)f(x)=√3sin(2X-π/6)+2sin²(x-π/12)=√3sin(2X-π/6)+1-cos(2x-π/6)=3sin(2X-π/6)-cos(2x-
函数f(x)=sinx的单增区间是【-π/2+2kπ,π/2+2kπ】所以,y=3sin(2x+π/3)的单增区间即,-π/2+2kπ
x+y=-3x=-y-3xy=2(-y-3)y=2-y²-3y-2=0y²+3y+2=0(y+1)(y+2)=0y1=-1y2=-2一、当y1=-1时,x1=-2,根号y分之x加根
解:(1)2kpi-pi/2
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因为,-π/2
x∈【0,π】π/6-2x∈[-11π/6,π/6] sinx在[-π3/2,-π/2]上是减函数,所以π/6-2x=-π3/2 x=5π/6π/6-2x=-π/2
设函数f(x)=cos(2x派/3)sin^2x?1:求f(x)的值域和最小正故值域为[1/2-根号3除以2,1/2根号3除以2],最小正周其为π
解答;f(x)=sin(2x+3分之π)∴sin(2x+π/3)=-3/5∵x∈(0,π/2)∴2x+π/3∈(π/3,4π/3)∵sin(2x+π/3)
1:y=2sin(x+π/6)-2cosx=2[sinxcosπ/6+cosxsinπ/6]-2cosx=√3sinx+cosx-2cosx=√3sinx-cosx=2sin(x-π/6)2:y=2c
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/
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+