nynw.net
当前位置:首页 >> 设函数F(x)=2根号3sinxCosx+2Cos^2x+m(1)求函数F... >>

设函数F(x)=2根号3sinxCosx+2Cos^2x+m(1)求函数F...

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(...

f(x)=√3sinxcosx-cos²x-1/2 =(√3/2)sin2x-(1+cos2x)/2-1/2 =(√3/2)sin2x-(1/2)cos2x-1 =sin(2x-π/6)-1. 早小正周期T=2π/2=π. sin(2x-π/6)=-1, 即x=kπ-π/6时, 最小值f(x)|min=-2。

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

f(x)=2√3sinxcosx+2cos^2x-1=√3sin2x+cos2x=2sin(2x+π/6) 由π/2+2kπ≤2x+π/6≤3π/2+2kπ得:π/6+kπ≤x≤2π/3+kπ,因此单调递减区间为 [π/6+kπ,2π/3+kπ] f(x)=2sin(2x+π/6)的最大值为2,此时2x+π/6=π/2+2kπ,即x=π/6+kπ,所以自变量x的集合为 {x|x=π/6...

f(x)=2√3sinxcosx+1-2(sinx)^2 =√3sin2x+cos2x =2sin(2x+π/6) 所以最小正周期=2π/2=π -π/2+2kπ

f(x)=(√3)[sin(x/2)][cos(x/2)]+cos²(x/2)? 是吗? 如果是的话: 最小正周期:2π/(1/2)=4π 单调递增区间是:x∈(4kπ-7π/6,4kπ+5π/6),其中:k∈N 解: f(x)=(√3)[sin(x/2)][cos(x/2)]+cos²(x/2) f(x)=cos(x/2)[(√3)sin(x/2)+cos(x/2)]...

解: 原式=1-(1+cos2x)/2+√3sin2x-1/2cos2x. x∈R. =√3sin2x-cos2x+1/2. =2[(√3/2sin2x-1/2cos2x+1/2. ∴f(x)=2sin(2x-π/6)+1/2. f(x)的最小证周期T=2π/2=π; 当sin(2x-π/6)=1时,f(x)max=2*1+1/2=5/2; 当sin(2x-π/6)=-1时,f(x)min=2*(-1)+1/2=-3...

解:f(x)=2cos²x+2√3sinxcosx =(2cos²x-1)+1+√3×2sinxcosx =cos2x+√3sin2x+1 =2sin(2x+π/6)+1 函数f(x)最小正周期T=2π/2=π 由2kπ≤2x+π/6≤2kπ+π/2 (k∈z),得 kπ-π/12≤x≤kπ+π/6 或 由2kπ+3π/2≤2x+π/6≤2kπ+2π(k∈z),得 kπ+2π/3≤x≤kπ+11π/...

网站首页 | 网站地图
All rights reserved Powered by www.nynw.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com