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已知函数F(x)=2根号3sinxCosx+2Cos²x.(1...

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

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]

解: (1) f(x)=2√3sinxcosx+2cos²x-1 =√3sin(2x)+cos(2x) =2[(√3/2)sin(2x)+(1/2)cos(2x)] =2sin(2x+π/6) 最小正周期T=2π/2=π (2) f(x0)=6/5 2sin(2x0+π/6)=6/5 sin(2x0+π/6)=3/5 x0∈[π/4,π/2] 2π/3≤2x0+π/6≤7π/6 cos(2x0+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

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)=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...

已知函数f(x)=2sin²x+2根号3sinxcosx+1,求函数的单调递增区间 函数在区间【0,п/2】上的最值 解析:∵函数f(x)=2sin²x+2√3sinxcosx+1=-cos2x+√3sin2x+2 =2sin(2x-π/6)+2 其单调增区间为: 2kπ-π/2

1. f(x)=√3sinxcosx-cos²x+1/2=(√3/2)(2sinxcosx)-(1/2)(2cos²x-1) 二倍角公式:2sinxcosx=sin(2x),2cos²x-1=cos(2x),于是有: f(x)=(√3/2)sin(2x)-(1/2)cos(2x), cos30°=cos(π/6)=√3/2,sin30°=sin(π/6)=1/2,于是有: f(x)=sin(2...

(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

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