2sin(x π 4 ) 3在R上恒成立,则t的取值范围是t≥ .
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f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x=[sinx+(√3)cosx+sinx]cosx-(√3)(sinx)^2=2sinxcosx+(√3)[(cosx)^2-(
(1)列表:2x+π30π2π3π22πx-π6π12π37π1256y020-20做出函数在一个周期上的简图,再根据图象的周期性特征,得到在一个周期[0,π]上的图象.(2)函数f(x)的最小正周期
f(x)=2sin^2(π/4+x)-√3cos2x-1=1-cos(π/2+2x)-√3cos2x-1=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2(1/2sin2x-√
中间是不是+?则y=-sin(3x-π/4)所以y递增则sin递减所以2kπ+π/2再问:第一步去分母的时候3边乘以4,左边和右边应该变成了8kπ了的呀再答:-3x+兀/4再问:??什么意思,-3x+
f(x)=1+sin2x+2cos^2x=1+sin2x+1+cos2x=√2sin(2x+π/4)+22kπ-π/2≤2x+π/4≤2kπ+π/2kπ-3π/8≤x≤kπ+π/8[0,π/8]
Tmin=2π/2=π2x+π/6∈(-π/2+2kπ,π/2+2kπ),k∈Zx∈(-π/3+kπ,π/6+kπ),k∈Z第二小题是甚?再问:⑵函数f(x)的图象可以由函数y=sin2x(x∈R)的
已知函数f(X)=sin^2wx+根号3sinwx*sin(wx+π/2)+2cos^2wx,x属于R,在y轴右侧的第一个最高点的横坐标为π/6,求w;若将函数f(x)的图像向右平移π/6个单位后,再
f(x)=sin(2x+7π/4)+cos(2x-3π/4)=2sin(2x-π/4)由-π/2+2kπ≤2x-π/4≤π/2+2kπ,k∈Z,得-π/8+kπ≤x≤3π/8+kπ,k∈Z,即f(x)
y=-sin(3x-π/4)递增则sin(3x-π/4)递减所以2kπ+π/2
当(2x-π/3)在(kπ,(k+1)π)时为递减吧k为整数当(-3x+π/4),在(-π/2+kπ,π/2+kπ)为递增然后自己解了
f(3a+π/2)=2sin(1/3(3a+π/2)-π/6)=2sina=10/13,sina=5/13,cosa=12/13f(3b+2π)=2sin(1/3(3b+2π)-π/6)=2cosb=
f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x
f(x)=sinx+sin(x+2π/3)=simx+simxcos2π/3+cosxsin2π/3=sinx-1/2sinx+√3/2cosx=1/2sinx+√3/2cosx=sin(x+π/3)
因为函数y=sinx在[-π/2+2kπ,π/2+2kπ],所以-π/2+2kπ《-3x+π/4《π/2+2kπ,得-3π/4+2kπ《-3x《π/4+2kπ,解得-π/12-2/3kπ《x《π/4-
(1)化解函数:√3∵f(x)=sin²wx+√3sinwxcoswx+2cos²wx=√3/2sin2wx+sin²wx+cos²wx+cos²wx
1)最小正周期T=2π/2=π取得最大值时x满足2x-π/4=π/2+2kπk属于正整数即x=3π/8+kπk属于正整数因此f(x)取得最大值时x的集合为{x|x=3π/8+kπk属于正整数}(2)分
f(x)=[2(sinx*1/2+cosx*√3/2)+sinx]cosx-√3sin²x=(2sinx+√3cosx)cosx-√3sin²x=2sinxcosx+√3(cos&
解(1)由f(x)=sin(π/3-2x)=-sin(2x-π/3)当2kπ-π/2≤2x-π/3≤2kπ+π/2,k属于Z时,y是减函数,即kπ-π/12≤x≤kπ+5π/12,k属于Z时,y是减函
令2x+π/4=3/2π+2kπ,k∈N,此时,函数取得最小值3*(-1)+1=-2取值范围,x=5/8π+kπ,k∈N