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设F(x)=2AsinxCosx+2BCos2x%B,(A,B∈R+)若F(...

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(1)f(x)=3sinx4cosx4+cos2x4=32sinx2+12cosx2+12=sin(x2+π6)+12,∵f(x)=1,∴sin(x2+π6)=12,∴x2+π6=2kπ+π6,或x2+π6=2kπ+5π6,∴x=4kπ,或x=4kπ+4π3,∴cos(2π3-x)=cos(2π3-4kπ)=cos

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

(1) f(x)= 2sinxcosx+2 cos 2 x 2cosx =sinx+cosx(cosx≠0) ,…(4分)由题意可得 f(x)= 2 sin(x+ π 4 ) =0,故 x+ π 4 =kπ,即 x=kπ- π 4 (k∈Z) . …(2分)(2)当 x∈[0, π 3 ] 时,方程 a=sinx+cosx= 2 sin(x+ π 4 ) 有两个不同解,等价于...

因为2sinx*cosx=sin2x,cos^2(x)-sin^2(x)=cos2x 所以f(x)=asinx-cosx

解 f(x)=a*b =√3cosxsinx-1/2cos2x =√3/2(2sinxcosx)-1/2cos2x =√3/2sin2x-1/2cos2x =sin2xcosπ/6-sinπ/6cos2x =sin(2x-π/6) 最小正周期为; T=2π/2=π ∵x∈[0,π/2] ∴2x-π/6∈[-π/6,5π/6] ∴ 当2x-π/6=-π/6时 f(x)取得最小值,f(x)=-1 当2x-π/6=π/2...

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

解答:解:(1)函数f(x)=3?cos2x2-4t?sinx2cosx2+2t2-6t=sin2x-2tsinx+2t2-6t+1 =(sinx-t)2+t2-6t+1.当t<-1时,g(t)=(-1-t)2+t2-6t+1=2t2-4t+2.当-1≤t≤1时,g(t)=t2-6t+1.当t>1时,g(t)=(1-t)2+t2-6t+1=2t2-8t+2.综上可得...

f(x)-g(x)=(1-a)cosx-(1+a)=0 (1-a)cosx-(1+a)=0 cosx=(1+a)/(1-a) 则 -10 a

a=(1/2,√3sinx) b=(cos2x,cosx) f(x)=a*b =(1/2)cos2x+√3sinxcosx =(1/2)cos2x+(√3/2)sin2x =sin(2x+π/6) =sin[2(x+π/12)] 剩下的应该会了吧 楼主~ 很高兴能帮助到你。若满意记得“采纳为满意答案”喔!祝你开心~O(∩_∩)O~

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