 In this video, we're asked to solve the equation sine of 2x times cosine of x plus cosine of 2x times sine of x equals root 2 over 2 and we're going to do this on the domain 0 to 2 pi. So we want to solve this in terms of radian measure. Okay, so you look on the left hand side, there's a lot going on here, right? We have a sine of 2x, a cosine of x. We have a cosine of 2x and a sine of x. What's going on here? We have these mismatched angles, 2x and x and we also have sines and cosines together. Perhaps there's some way of simplifying this is there's some type of identity that could be useful here. And so we start searching like this is a product of some this is some of the product. It turns out it's much easier than either of those. This is going to be the angle some identity worked in reverse, right? When you look at the sine of 2x cosine of x plus cosine of 2x times sine of x. This looks a lot like the angle some identity for sine. The left hand side, which again, the reminder here, if you take sine of a plus b, this is equal to sine of a cosine of b plus cosine of a sine of b. So if we set a equal to 2x and b equal to x, this is exactly what we need here. So the left hand side would simplify just to be sine of 2x plus x and then the right hand side is root 2 over 2. Holy cow, that was so much easier if you use the right identity. The left hand side then simplifies to be sine of 3x and this is equal to root 2 over 2. When is sine equal to root 2 over 2? That sounds like a 45 degree angle. Whoops, we're in radians. That should be pi force. So we're going to get 3x equals pi force plus 2 pi k. That's the solution in the first quadrant. Sine is also positive in the second quadrant. So we need the English references to pi force in the second quadrant. Just set it at 3 pi force plus 2 pi k, like so, in which case then to solve for x, we would divide both sides by 3, like so. And so we get the solution x equals pi twelfths plus 2 pi thirds k. And then we get, in this case, you're going to take 3 pi fourth divided by 3, which you can say 3 pi twelfths, but that's just going to be pi force, like so. And then you're going to get 2 pi thirds k. 2 pi thirds, of course, is the new period. No sine of 3x there. It's periods now got cut in by a third, like so. Now, we're not quite done yet. This is the general solution, but we're asking for what's the solution between zero and 2 pi. So we actually need to list each and every one of them out. So what are the solutions here? We're going to get x equals looking at the first general solution. You get pi twelfths. You would also get pi twelfths plus 2 pi thirds. You would also get pi twelfths plus 4 pi thirds. Where do the 4 pi thirds come from? They come from 2 times 2 pi thirds. Because after all, this k here is an integer. There's some options. So we took, for example, when k equals 0, you get pi twelfths. When k equals 1, you're going to get just 2 pi thirds. When k equals 2, you're going to get 4 pi thirds. And so as you go from 0 to 2 pi, 2 pi thirds, you could iterate through that 3 times going from 0, 1, and 2. Alright, we have to do that for the other possibility as well. We're going to take pi fourths plus 0, so just pi fourths. You're going to get pi fourths plus 2 pi thirds. And then you're going to get pi fourths plus 4 pi thirds, which again, like we saw before, this is k equals 0. This is k equals 1. This is k equals 2. And so it would be good form to add these things together. And if you could put them in order, that's also helpful here. The smallest of which is going to be the pi twelfths. The next of which would be the pi fourths. And you'll notice that this little group, this is the group that corresponds to k equals 0, of course. What else could you do? Well, there's a group for the k equals 1. So what about those? So if you look at k equals 1, there's this one right here. You have pi over 4 plus 2 pi over 3. And you're going to have pi over 12 plus 2 pi over 3. What's going to happen in that situation? We have to add these things together, of course. So if you consider the first one right there, in order to add this together with 12, we need the denominator to be 12 as well. We're actually going to times the bottom by 4, and then hence the top by 4 as well. So you're going to get 4 times 2, which is 8. 8 pi plus pi gives us a 9 pi. So we have 9 pi over 12. But I guess we could simplify that. 9 and 12 have a common factor of 3. So we could simplify that down to be just a 3 pi over 4. Like so. Then for the next one, we have to basically do the same thing. We need a 4 times 4 here to give us the 8 pi over 12 again. But we need a 3 times 3 right here. So you're going to get 3 plus 8, which is 11. So you get 11 pi over 12. Like so. This is the cluster associated to k equals 1. And then the last one here, we're going to get those associated to k equals 2. Looking at this one right here, again change the denominator. We're going to times the bottom by 4, the top by 4. So we're going to get 4 times 4, which is 16. 16 pi plus 1 is 17. So we end up with a 17 pi over 12. And then the next one, same basic idea, you take the top and bottom times it by 4 again. And then this one, you'll times it by 3. In which case, then you're going to get 3 pi plus 16 pi, which is a 19 pi, like so over 12. And so these are the solutions that coincide with k equals 2. This gives us all 6 solutions where 6 come from. Well, sine had 2 general solutions right here. But since we cut the period, those general solutions become 3 times as more frequent. So 2 times 3 gives us 6 solutions total.