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3Cos2x sin2x

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...

如图

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.

(1)f(X)=sin2X+√3cos2X=2(1/2sin2X+√3/2cos2X)=2(cosπ/3sin2x+sinπ/3cos2x) =2sin(2x+π/3) 即f(x)=2sin(2x+π/3) (2)∵-1≤sin(2x+π/3)≤1 ∴-2≤2sin(2x+π/3)≤2 ∴函数的最大值为2,最小值为-2 (3)最小正周期为T=2π/2=π (4)令2x+π/3=-π...

∵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...

D

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

y = 10*[(√3/2)*cos(2x) +(1/2)*sin(2x)] = 10*[sin(π/3)*cos(2x) + cos(π/3)*sin(2x)] = 10sin(π/3+2x) 所以,最小正周期: T = 2π/2 = π

楼主少了括号哦,老师这样写是会被扣分滴哦~等我一下我上图,求不忙采纳其他人的。。。

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