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3Cos2x sin2x

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f(x)=sin2x+3cos2x=2(12sin2x+32cos2x)=2sin(2x+π3).∵f(x)的图象关于点(x0,0)成中心对称,∴f(x0)=0,即2sin(2x0+π3)=0,∴2x0+π3=kπ,x0=kπ2?π6,k∈Z,∵x0∈[0,π2],∴x0=π3.故选:C.

∵f(x)=sin2x-2sinxcosx+3cos2x=1-2sinxcosx+2cos2x=1+1+cos2x-sin2x=-(sin2x-cos2x)+2=-2sin(2x-π4)+2.∴把函数f(x)的图象沿x轴向左平移m(m>0)个单位,得到函数g(x)的图象的解析式为:g(x)=-2sin(2x+2m-π4)+2.∵函数g(x)的图象关...

先提2,然后令cos阿尔法等于二分之一就行了

特征方程为r²+1=0,得r=i, -i 设特解为y*=acos2x+bsin2x y*'=-2asin2x+2bcos2x y*"=-4acos2x-4bsin2x 代入方程得:-4acos2x-4bsin2x+acos2x+bsin2x=3cos2x+sin2x -3acos2x-3bsinx=3cos2x+sin2x 比较系数得:-3a=3, -3b=1 得a=-1, b=-1/3 因...

f(x)=sin(2x+π/3)+√3/3sin2x-√3/3cos2x =1/2sin2x+√3/2cos2x+√3/3sin2x-√3/3cos2x =( 1/2+√3/3)sin2x+√3/6cos2x =√(2/3+√3/3)sin(2x+φ) tanφ=√3/6/( 1/2+√3/3)=2-√3 φ=π/12 ∴fx的最小正周期=2π/2=π sinx的对称轴:x=kπ+π/2 ∴2x+π/12=kπ+π/2 x=kπ...

3cos2x+4sin2x = 5[ (3/5)cos2x + (4/5)sin2x] =5cos(2x- arccos(3/5)) ∫dx/(3cos2x+4sin2x) =(1/5)∫dx/cos(2x- arccos(3/5)) =(1/5)∫sec(2x- arccos(3/5)) dx =(1/10)ln|sec(2x- arccos(3/5)) + tan(2x- arccos(3/5))| + C

(1)∵tanx=3 ∴sinx=3cosx 又sin²x+cos²x=1 ∴(3cosx)²+cos²x=1 ∴cos²x=1/10 ∴sinxcosx=3cos²x=3/10 (2)y=sin2x-√3cos2x=2[sin2xcosπ/3-cos2xsinπ/3]=2sin(2x-π/3) 当2x-π/3∈[2kπ-π/2,2kπ+π/2]即x∈[kπ-π/12,kπ+5π/...

(Ⅰ)f(x)=3cos2x+2cosxsinx+sin2x=3×1+cos2x2+sin2x+1?cos2x2=2+sin2x+cos2x=2sin(2x+π4)+2当2x+π4=π2+2kπ,即x=kπ+π8(k∈Z)时,f(x)取得最大值2+2;(Ⅱ)当?π2+2kπ≤2x+π4≤π2+2kπ,即kπ?3π8≤x≤kπ+π8(k∈Z)时,正弦函数sin(2x+π4)单调...

2×(1/2cos2x-√3/2sin2x) =2(cosπ/3cos2x-sinπ/3sin2x) =2cos(2x+π/3)

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