i think there are 4 answers to this , but im not sure how to do this.....
Do you mean
$\displaystyle 2\sin^2{x}=1\text{ for }(0<x<2\pi)$ ?
If so
Solve the equation, and then restrict the infinite amount of solutions to the given interval. It's lookin like $\displaystyle \sin{x}=\pm\frac{\sqrt{2}}{2}$
Last edited by VonNemo19; Nov 7th 2009 at 04:13 PM.