# Thread: Solving a difficult equation

1. ## Solving a difficult equation

without using a graphing tool how do you solve $\displaystyle cos(t)=e^{-2t}$...heres as far as I got $\displaystyle \frac{e^{it}+e^{-it}}{2}=e^{-2t}$...then $\displaystyle e^{it}+e^{-it}=2e^{-2t}$...so $\displaystyle e^{it}+e^{-it}=e^{\ln(2)-2t}$...here is where I don't know where to go...anybody have any ideas?

2. Hello,

What if you use infinite series ?

3. Originally Posted by Mathstud28
without using a graphing tool how do you solve $\displaystyle cos(t)=e^{-2t}$...heres as far as I got $\displaystyle \frac{e^{it}+e^{-it}}{2}=e^{-2t}$...then $\displaystyle e^{it}+e^{-it}=2e^{-2t}$...so $\displaystyle e^{it}+e^{-it}=e^{\ln(2)-2t}$...here is where I don't know where to go...anybody have any ideas?
There are no exact solutions to this equation so it is not surprising that you haven't been able to figure out how to solve it.

-Dan

4. Originally Posted by Mathstud28
without using a graphing tool how do you solve $\displaystyle cos(t)=e^{-2t}$
[snip]
Why? Curiosity?

5. In general math equations have no closed form solutions. But that does not bother mathematicians. They only care about whether a solution exists.

6. ## Why?

The reason why is that in my calc class which I have already finished and passed the next two classes but am forced to take anyways I am the guy that knows everything and can do things in a second that noone else has any idea how to do....so one day we were doing something and we neede to solve this...and one kid goes..."Hey how do you solve this" to me...and I go well its simple you just use the definition...and I did what I showed you guys...then I lost steam...I thought I could do it but apparently it can't be done...thanks for your response!

7. Is this pretention ?

Well, you can use different methods of approximation. For equations including exp and log, i don't really know, but perhaps you can find it somewhere

8. ## No this is not

pretension...I was jsut merely explaining the circumstance and why I felt as though I had to get an answer due to a feeling of obligatory need to perform...and thanks I will find an approximation...or just use a calculator!