I need help with this Precal problem

find all the exact t- values for which sin t = -1/2
I need to show work also so any steps would be helpful!!

We're not supposed to DO your work. I'll therefore give you an example instead.

If you had cos x = 1, then

\alpha = cos^{-1} (1) = 90

And the answers would be

| /
S | A
____|/____
|\
T | C
| \

In 'A' and 'C', ( meaning all positive and cos positive)

First angle is 90 degrees, you get it directly.
The second is 270, because 360 - 90 = 270 degrees.

Therefore, x would equal 90 and 270 degrees. (note that the angle you have is between the diagonal and the horizontal line)
---------

If you had cos x = -1, then

\alpha = cos^{-1} (1) = 90

And the answers would be

\ |
S | A
___\|_____
/|
T | C
/ |

In 'S' and 'T', ( meaning tan positive and sine positive) since you had cos x = -1 and if it is negative, it cannot be all positive nor cos positive.

First angle is 90 degrees, you get it by 180 - 90 = 90 degrees
The second is 270, because 180 + 90 = 270 degrees.

Therefore, x would equal 90 and 270 degrees.
-----------

If you had sin x = 0, then

\alpha = sin^{-1} (0) = 0

And the answers would be

\ | /
S | A
___\|/____
|
T | C
|

In 'A' and 'S', ( meaning all positive and sine positive)

First angle is 0 degrees, you get it directly.
The second is 180, because 180 - 0 = 180 degrees.
The third is 360 because 360 - 0 = 360 degrees.

Therefore, x would equal 0, 90 and 360 degrees.
----------

I know it's a bit confusing at first, so don't hesitate to outline the parts you don't understand.