calculus

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

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  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.

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  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?

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  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?

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

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