Hello from MrBillDoesMath!
Answer:
pi/2, - 3pi/2
Discussion:
The solutions of sin(t) = 1 are (pi/2) + 2*pi*n, where n = 0, -1,1, -2,2,-3,3, etc.
In particular, sin(pi/2) = 1 ( n = 0) and sin(-3pi/2) -1 ( n = -1) and those argument values lie in the interval [-2pi, 2pi]
Regards,
MrB
P.S. I'll be on vacation from Friday, Dec 22 to Jan 2, 2019. Have a Great New Year!
