corosync crm 配置:求2cos(x/2)-3sin(x/3)周期

来源：百度文库编辑：高考问答时间：2024/04/27 21:45:50

答案是6∏，请告诉我过程

2cos(x/2)-3sin(x/3)
令x=x+T
2cos((x+T)/2)-3sin((x+t)/3)
=2(cos(x/2)cos(T/2)-sin(x/2)sin(T/2))-3(sin(x/3)cos(T/3)+cos(x/3)sin(T/3))...............(1)

若T为周期,则原式和(1)式应该相等.
即:
cos(T/2)=1.....................(a)
sin(T/2)=0.....................(b)
cos(T/3)=1.....................(c)
sin(T/3)=0.....................(d)

由(a)(b):T/2=∏+k∏
=> T=2∏+2k∏ (k 为使得T为正数的整数,下同)
由(c)(d):T/3=∏+k'∏
=> T=3∏+3k'∏
因此当k=2,k'=1时T=6∏

2cos(x/2) 周期4∏
3sin(x/3) 周期6∏
最小公倍数为12∏

答案：12∏

求Lim (2sin^n x+3cos^n x)∕(sin^n x+cos^n x) ,0≤x≤π/2 求2cos(x/2)-3sin(x/3)周期三角函数题目急求:2cos(X/2)-3sin(X/3) 求cos（x/2)+2*sin(x/3)的周期求函数y=sin平方X+2sin xcos x+3cos平方 6sin平方x+sinxcosx-2cos平方x=0,x属于[TT/2,TT]，求sin(2x+TT/3) 已知(4sin^2) x-(cos^2)x-6sinx+3cosx=0,求(cos2x-cos^2 x)/(1-cot^2 x)的值已知(4sin^2) x-(cos^2)x-6sinx+3cosx=0,求(cos2x-cos^2 x)/(1-cot^2 x)的值已知(4sin^2) x-(cos^2)x-6sinx+3cosx=0,求(cos2x-cos^2 x)/(1-cot^2 x)的值。已知(4sin^2) x-(cos^2)x-6sinx+3cosx=0,求(cos2x-cos^2 x)/(1-cot^2 x)的值。