Asked by Gordon
How to solve 2.5e^-t cos2t=0?
Answers
Answered by
Steve
you have the product of three factors
(2.5)(e^-t)(cos 2t) = 0
2.5 is never zero
e^-t is never zero
cos 2t = 0 when t = (2k+1)(pi/4) for integer k.
(2.5)(e^-t)(cos 2t) = 0
2.5 is never zero
e^-t is never zero
cos 2t = 0 when t = (2k+1)(pi/4) for integer k.
Answered by
Gordon
But I don't understand how t = (2k+1)(pi/4)?
From what I calculated, 2t= cos^-1 0
t= 45
Is it correct?
From what I calculated, 2t= cos^-1 0
t= 45
Is it correct?
Answered by
Steve
you do know that cos(z) = 0 when z is an odd multiple of pi/2, right?
So, cos(2t) = 0 when 2t is an odd multiple of pi/2; that is, t is an odd multiple of pi/4.
45 degrees is an odd multiple of pi/4, right?
So, cos(2t) = 0 when 2t is an odd multiple of pi/2; that is, t is an odd multiple of pi/4.
45 degrees is an odd multiple of pi/4, right?
Answered by
Gordon
Now I get it..thanks^^ really helps a lot!!