When doing this kind of problem we need to think about when is cosine 1. So cosine is one at the angle 0. So this cosine is one at 0. So if it had been cosine X equal 1, we would have said X = 0 + 2 K π because of the period it repeats. But it wasn't cosine of X, it was cosine of this two Theta minus π halves. So what we're going to say is this 2 Theta minus π halves is going to equal 0 from that's when cosine is one. But it happens every two K π. So then we're going to solve for Theta zero plus anything is itself. So we're going to add the Pi halves over and then we're going to divide by two. So we get Theta as Pi 4 + K Pi. Now the directions say we want the answers in between 0 and 2π. So we think about what happens if K equal 0. If K is zero, we get Theta equaling π force plus 0 * π or just π force. If K equal 1, we get Theta equaling π force plus 1 * π, which is five Pi force because Pi is 4 / 4 Pi. If we thought about what if K equal 2? If K is 2, we're going to be too much because π force +2 Pi 2π is 8 Pi force plus a π force, which would be 9 Pi force. That's too big. Well, what if K had been -1? If K is -1, we'd get π force plus -1 * π and that would give us -3 Pi force, and that would be too small because we needed it to be 0. So our answers are going to be this π force and five Pi force. Once again, we have to have it equaling 1. Figuring out what that angle is, and then realize that the angle here wasn't just a plain Theta or X, it was this whole parenthesis. And we're solving. Now a nice little easy way to help is the fact that it was two times Theta meant that I should expect twice as many answers as what I usually get in a full. And if we think about our cosine being one in a full. That only occurs one time, so right there at that height of 0 or at 2π. So we get 2 answers because two times Theta.