[riverakathy]

On the problem: 1/1-sin(x), wouldn't the domain just be x != pi over 2, since sin of 90 degrees is pi over 2, and what would be 1-1 in the denominator? The book says somthing like: x!= (2n + .5)pi n= 0,+-1, +-2 but I don't understand that solution. Anyone know? Thanks!

[Qxsumehj]

The answer is cyclical so if no range is specified you need to define the whole domain. That solution specifies all points where sin(x) = 1.

[tramdoctorsss]

The answer is cyclical so if no range is specified you need to define the whole domain. That solution specifies all points where sin(x) = 1.

Shouldn't the domain be every number where X is defined? In other words, the domain will be every possible x value except for values in which sin(x) is NOT equal to 1? since that would be 1/0 = undefined.

[Qxsumehj]

Yes, he's specified that the solution in the book says x is defined everywhere except those points, hence the "!=" rather than "=" and he doesn't understand why. I've explained why there is an extension to his answer, but not explicitly put it into the same conext, I've just followed as to why there is more than one solution to sin(x) = 1.