Recent content by Shaggy16

Case 2 is correct, but in case 1, you should have added 1 instead of subtracting it. Case 1: cos(2x)=1/2 2cos^2(2x)-1=1/2 2cos^2(2x)=1/2+1=3/2 Try it from here. And you have to remember that the function is periodic, so there will be multiple solutions to each of those.

That's one answer, but there are more. Try the substitution u=cos(2x). That will give you a quadratic, which will give you both of the answers you need.

Okay, that was a really stupid question. An open set is just a set contained in the topology. If you are considering R^n, then it is implied that an open set is one such that at every point one could construct an open ball such that all of the points in the open ball are also in the set...

Then where do open sets come into play?

Homework Statement I started studying point-set topology a while ago, and I started to wonder, "Does a set have to be partially ordered in order to define a topology on it?" Homework Equations The Attempt at a Solution I know that every set in a topology has to be open, which...

I forgot all about the half-angle identities... Thank you

Homework Statement Is there a way to determine the period of a function like f(x) = a*sin(b*x)^2 + c*cos(d*x)^2 + e*sin(f*x) + g*cos(h*x)? Homework Equations The Attempt at a Solution I know how to find the periods of sines, cosines, and arbitrary sums of the two, but the...