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已知函数F(x)=2根号3sinxCosx+2Cos²x.(1...

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)=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+2cos²x =√3sin2x+cos2x+1 =2sin(2x+π/6)+1 f(4π/3)=2sin(8π/3+π/6)+1=2sin5π/6+1=2 x∈[0,π/2] 2x+π/6∈[π/6,2π/3] f(x)的值域:[2,3]

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

你题目的第一项与第三项, 可以改化为: cos2x, 第二项可以改写为 (√3)×sin2x, 于是, f(x)=2×sin{(2x)+(π/6)}, 然后自己可以完成任务啦! ——题目的a是啥意思?

(1) f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6) 函数的最小正周期T=π (2) f(x)=2sin(2x+π/6)=6/5 sin(2x+π/6)=3/5 x∈[π/4,π/2], cos(2x+π/6)=-4/5 cos2x=cos(2x+π/6-π/6)=cos(2x+π/6)cosπ/6...

(1)f(x)=(1-cos2x)/2+√3/2*sin2x+1+cos2x=√3/2*sin2x+1/2*cos2x+3/2=sin(2x+π/6)+3/2所以f(x)的最小正周期为π在(kπ-π/3,kπ+π/6)上单调递增,在(kπ+π/6,kπ+2π/3)上单调递减(2)f(x)可以由函数y=sin2x的图像向左平移π/12,向上平移3/2得到

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...

解由f(x)=√3/2sin2x+1/2(1-cos2x)+1/2 =sin(2x-π/3)+1 由x属于[-π/12,5π/12] 则2x属于[-π/6,5π/6] 则2x-π/3属于[-π/2,π/2] 则sin(2x-π/3)属于[-1,1] 则sin(2x-π/3)+1属于[-2,2] 故函数的值域为[-2,2]

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