Arccos problems

[Loki123]

So usually when I would use arccos, arcsin I'd have two answers. One that's just arccos x and one that is pi - arccos x. In this case the solution only had one. Why?

 

[Harry_the_cat]

What if you had \(\displaystyle \cos x =\frac{1}{2}\), what would your solution be?

Could you post the whole question please?

 

[blamocur]

[imath]x=\arccos a[/imath] is a solution of the [imath]\cos x = a[/imath] equation, but [imath]\pi - \arccos a[/imath] is not -- you might be confusing it with [imath]\arcsin[/imath] case.

 

[Deleted member 4993]

So usually when I would use arccos, arcsin I'd have two answers. One that's just arccos x and one that is pi - arccos x. In this case the solution only had one. Why? View attachment 30976

Was there a specified domain?

The second answer could be:

x = 2*pi - (3-y)/(2*y)