Answer
The t-intercepts of the graph of $~~y_1~~$ are $~~t = \pi~n,~~$ where $n$ is an integer
The x-intercepts of the graph of $~~y=sin~x~~$ are $~~x = \pi~n,~~$ where $n$ is an integer
The t-intercepts of $y_1$ are the same as the x-intercepts of the graph of $y = sin~x$
Work Step by Step
$y_1 = e^{-t}~sin~t$
Note that $~~e^{-t} \gt 0~~$ for all real numbers.
To find the t-intercepts of $y_1$, we can let $y_1 = 0$:
$y_1 = e^{-t}~sin~t = 0$
$e^{-t}~sin~t = 0$
$sin~t = 0$
$t = \pi~n,~~$ where $n$ is an integer
The t-intercepts of the graph of $~~y_1~~$ are $~~t = \pi~n,~~$ where $n$ is an integer
The x-intercepts of the graph of $~~y=sin~x~~$ are $~~x = \pi~n,~~$ where $n$ is an integer
Therefore, the t-intercepts of $y_1$ are the same as the x-intercepts of the graph of $y = sin~x$
