Question

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

Answer 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

 P.S.  I'll be on vacation from Friday, Dec 22 to Jan 2, 2019. Have a Great New Year!