Can anyone please explain to me how to solve sin(x)= -1 where the possible solutions for the angle is between 0 and 4pi ? i.e not in between 0 and 360 degrees?
x = arc sin (-1 )
Originally Posted by SuchAPirate
x = arc sin (-1 )
Doesn't that just give the reference angle though? which would only work for angles between 0 and 2pi?
i drew the graph out, so i got 2 solutions X1 = arcsin (-1) + pi and X2 = arcsin (-1) +3pi
so X1 = 3pi/2 and X2 = 7pi/2
Originally Posted by SirusFPS
Can anyone please explain to me how to solve sin(x)= -1 where the possible solutions for the angle is between 0 and 4pi ? i.e not in between 0 and 360 degrees?
sin(x) = -1 if x = 2pi*n + (3/2)*pi, where n = 0, 1, 2, 3, ....
so in the range of 0 to 4pi, your only answers here are (3/2)*pi and (7/2)*pi.
Originally Posted by Hero
sin(x) = -1 if x = 2pi*n + (3/2)*pi, where n = 0, 1, 2, 3, ....
so in the range of 0 to 4pi, your only answers here are (3/2)*pi and (7/2)*pi.
thank you, that makes sense to me!
Originally Posted by SirusFPS
thank you, that makes sense to me!
np, if you are a bit confused still or on anything else math related, post a message on my profile.