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?