已知函式f x 2 3sinxcosx 3sinx cosx

時間 2022-04-05 15:10:05

1樓:愛吃魚與魚

f(b)=1 b=60, a=30 c=90

已知函式f(x)=2√3sinxcosx-2cos²x+1 問函式的最小正週期和單調遞增區間?

2樓:墨書香閣

f(x)=2√3sinxcosx+1-2(sinx)^2=√3sin2x+cos2x=2sin(2x+π/6)所以最小正週期=2π/2=π-π/2+2kπ<=2x+π/6<=π/2+2kπ時遞增,所以遞增區間為[-π/3+kπ,π/6+kπ]

∵f(x)=sinxcosx+sin²x=0.5sin2x+0.5(1-cos²x)=0.5√2sin(2x-π/4)+0.5

∴最小正週期t=2π/2=π

∴當2x-π/4∈[-π/2+2kπ,π/2+2kπ](k∈z)時f(x)單調遞增

解得x∈[-π/8+kπ,3π/8+kπ],(k∈z)∴f(x)單調遞增區間[-π/8+kπ,3π/8+kπ],(k∈z)

已知函式f(x)=2√3sinx×cosx+2cos²x-1(x∈r) 10

3樓:韓增民鬆

已知函式f(x)=2√3sinx×cosx+2cos²x-1(x∈r)

<1>求函式f(x)的最小正週期及在區間[0,π/2]上的最值

<2>若f(x0)=6/5,x0∈[π/4,π/2],求cos2x0的值

(1)解析:∵函式f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2sin(2x+π/6)

∴f(x)的最小正週期為π

f(0)=2sin(π/6)=1,f(π/2)=2sin(π+π/6)=-1,f(π/6)=2sin(π/3+π/6)=2

在區間[0,π/2]上的最大值為2,最小值為-1

(2)解析:∵f(x0)=2sin(2x0+π/6)=6/5==>sin(2x0+π/6)=3/5

∴sin2x0*√3/2+cos2x0*1/2=3/5

√3sin2x0+cos2x0=6/5

sin2x0=2√3/5-√3/3cos2x0==> (sin2x0)^2=(2√3/5-√3/3cos2x0)^2=12/25-4/5(cos2x0)+1/3(cos2x0)^2

代入(sin2x0)^2+(cos2x0)^2=1

4/3(cos2x0)^2-4/5(cos2x0)^2-13/25=0

解得cos2x0=3/10-2√3/5或cos2x0=3/10+2√3/5

4樓:

簡化為f=√3sin2x+cos2x=2sin(2x+π/6)所以f(x)的最小正週期=2π/2=π

2x+π/6 ∈(π/6,7π/6)  所以 最大值為2   最小值為-1

f(x0)=6/5 =2sin(2x0+π/6),sin(2x0+π/6)=3/5,觀察 2x0+π/6 為鈍角 觀察下圖

設x   現有餘弦定理求直角邊為之長度  然後用  勾股定理  算出x   最後cos2x0=-40/x

已知函式f(x)=sin²x+2根號3sinxcosx-cos²x

5樓:小百合

f(x)=sin²x+2√3sinxcosx-cos²x=2√3sinxcosx-(cos²x-sin²x)=√3sin(2x)-cos(2x)

=2sin(2x-π/6)

1)2kπ-π/2≤2x-π/6<2kπ+π/2kπ-π/6≤x

2kπ+π/2≤2x-π/6<2kπ+3π/2kπ+π/3≤x

單增區間:[kπ-π/6,kπ+π/3]

單減區間:[kπ+π/3,kπ+5π/6]2)f(x)最大值為2

sin(2x-π/6)=1

2x-π/6=2kπ+π/2

x=kπ+π/3

已知函式f(x)=2√3sinx·cosx+cos^2x-sin^2x-1(x∈r)

6樓:匿名使用者

答:f(x)=2√3sinxcosx+cos²x-sin²x-1=√3sin2x+cos2x-1

=2*[sinx(√3/2)+cos2x(1/2)-1=2sin(x+π/6)-1

(1)f(x)=2sin(x+π/6)-1的單調增區間滿足:-π/2+2kπ<=x+π/6<=π/2+2kπ

所以:函式的單調增區間為[2kπ-2π/3,2kπ+π/3],k∈z(2)若-5π/12<=x<=π/3

則:-π/4<=x+π/6<=π/2

所以:-√2/2<=sin(x+π/6)<=1所以:-√2<=2sin(x+π/6)<=2所以:-1-√2<=f(x)<=1

所以:f(x)的值域為[-1-√2,1]

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