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

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.

But I don't understand how t = (2k+1)(pi/4)?

From what I calculated, 2t= cos^-1 0
t= 45

Is it correct?

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?

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

To solve the equation 2.5e^(-t) cos(2t) = 0, we can use the zero product property. According to the zero product property, if the product of two numbers is equal to zero, then at least one of the numbers must be zero.

First, let's set each factor equal to zero individually:

Setting 2.5e^(-t) equal to zero:
2.5e^(-t) = 0

Since e^(-t) can never be zero for any value of t, this equation has no solutions.

Now, let's set cos(2t) equal to zero:
cos(2t) = 0

To find the values of t that satisfy this equation, we need to find the values at which the cosine function equals zero. The cosine function equals zero at multiples of pi/2.

Therefore, we can write:

2t = n(pi/2), where n is an integer representing the number of complete cycles of the cosine function.

To solve for t, divide both sides of the equation by 2:

t = n(pi/4), where n is an integer.

So the general solution for the equation 2.5e^(-t) cos(2t) = 0 is t = n(pi/4), where n is an integer. This means that any value of t that is a multiple of pi/4 will satisfy the equation.