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已知函数F(x)=2根号3sinXCosX+2Cos^2X%1(X属于R)

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根号3sinxcosx+2cos^2x-1 =√3 sin2x+cos2x =2sin(2x+π/6) f(A)=2sin(2A+π/6)=2/3 sin(2A+π/6)=1/3 0

解:由已知条件变形得:f(x)=√3sin2x+cos2x, x∈R. f(x)=2sin(2x+π/6). ∵x∈[0,π/2], ∴2x+π/6∈[π/6,π+π/6]. ∵f(x)在[0,π/6]区间为增函数,在[π/6,π/2]区间为减函数,∴f(x)在x=π/6处取得最大值2; ∵f(π/2)<f(0),∴f(x)在x=π/2处,即2x+π/6=π+π/6处...

f(x)=2√3sinxcosx+2cos²x+m=√3sin2x+1+cos2x +m=2sin(2x+π/6) +m+1。 (1)f(x)的最小正周期为T=π。 由2kπ - π/2 ≤2x+π/6≤2kπ + π/2,得kπ - π/3 ≤x≤kπ + π/6 (k∈Z), ∴f(x)的单调递增区间为 [kπ - π/3,kπ + π/6](k∈Z)。 (2) 当x∈[0,π/6]时...

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(...

求采纳,

解: f(x)=2√3sinxcosx+1 -2sin²x =√3sin(2x)+cos(2x) =2[(√3/2)sin(2x)+(1/2)cos(2x)] =2sin(2x+π/6) -1≤sin(2x+π/6)≤1 -2≤2sin(2x+π/6)≤2 -2≤f(x)≤2 函数的值域为[-2,2]

f(x)=√3sinxcosx-1/2cos2x =√3/2sin2x-1/2cos2x =sin(2x-π/6). (1)最小值:f(x)|min=-1, 此时2x-π/6=2kπ-π/2, 即x=kπ-π/6 (k为整数); 最小正周期:T=2π/2=π. (2)f(C)=1,则 sin(2C-π/6)=1,即C=π/3. R=c/(2sinC)=√3/(2·√3/2)=1 (正...

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。

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