For what values of t on the interval [-2pi, 2pi] is sint=1?

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

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