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!!
    
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