a)t=3 b)t= 6 c)t=-3 d)=-6. Explain how you got your answer. I just wanna know how i would do this on the unit circle since it is not pi.
the numbers above are regular whole numbers I think no pi
If t is in radians, since 3 is > π/2 but < π, it is in quadrant 2 so subtract it from π: π - 3
6 is between 3π/2 and 2π so it's Q4 so subtract from 2π: 2π - 6
-3 would be the other way around so in Q3; add it to π: π - 3
and -6 is the other way around too so in Q1; add to 2π: 2π - 6
Maybe anyway, see what you think