Math problem

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?

That is the only way to solve it @SirusFPS

Originally Posted by SuchAPirate

That is the only way to solve it @SirusFPS

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.

Math problem

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