How to solve 2.5e^-t cos2t=0?

Answer 1

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.

Answer 2

But I don't understand how t = (2k+1)(pi/4)?
From what I calculated, 2t= cos^-1 0
t= 45

Is it correct?

Answer 3

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?

Answer 4

Now I get it..thanks^^ really helps a lot!!