 In a previous video, we attempted to solve the equation sine theta minus cosine theta equals one, and we wanted to find all solutions from zero to two pi. And we were successful in doing that. The basic idea is we move cosine to the side and we squared both sides. We got sine squared equals one plus cosine quantity squared, you foiled that out. And then we use a Pythagorean identity to turn into a quadratic equation, we solve it. I wanna revisit this problem here because when we've saw it in the previous lecture, we weren't considering any changes to the period. If you allow for period changes with your trigonometric functions, it turns out there's other ways you could solve these various problems and then maybe you like it better. So let's see a different approach to this one. This time I'm not gonna separate the trigonometric functions because after all, unless you were trained to do so, you might not think of doing such a thing. So let's just leave it as it is and just square both sides. Now, noticed by squaring both sides, we are potentially entering into the equation, party crash or solutions, which are not really solutions, they're fakers. So we have to make sure we check for those at the end of this problem. On the left-hand side, if we foil, we end up with sine squared theta minus two sine theta, cosine theta, and then we get a plus cosine squared theta like so on the right-hand side, we get one. Now, we're still gonna use a Pythagorean identity. We end up with of course here a sine squared plus cosine squared. That's gonna give us, of course, one. We just end up with a one minus two sine theta, cosine theta, this equals one. We can subtract one from both sides. We end up with negative two sine theta, cosine theta is equal to zero. Now, at this moment, you could be like, okay, sine equals zero, of course, at zero and pi. Cosine equals zero at pi halves and three pi. And then you could proceed to eliminate who's a party crasher from there. We'll come back to that in a second. This is the main, and that would be a perfectly valid approach, something that would be very appropriate for the previous lecture. What I wanna do instead, though, is look at this. Well, we have a two sine theta, cosine theta. That's just negative sine of two theta equals zero. Times both sides by negative one, you get sine of two theta equals zero, for which then we get that two theta is gonna equal, well, what is sine equal to zero? Sine equals zero at zero, so you take any two pi multiple of that, or sine of zero at pi, so you're gonna get pi plus two pi k as well. So really, if you have even multiples, if you have zero plus two pi k and then pi plus two pi k, those are just all multiples of pi, so you could just simplify that just as pi k, right? Like so, and then theta, if you divide by two here, two theta divided by two, of course, is theta. Pi k divided by two is you're gonna get pi halves k. And notice then you reproduce all of these answers here again. So you're gonna get, if you take k to be zero, you get zero. If you take k to be one, you're gonna get pi halves. If you take k to be two, you get pi. If you take k to be three, you get three pi halves, and then at four, you've wrapped around. So you get all the same solutions, and these is what we saw previously when we tried this video another time. Now, of course, you do have to check these solutions, of course. If you plug in zero, you're going to get zero, excuse me, you're gonna get zero minus one, which is negative one, that doesn't work. So zero, we take away. If you try pi, you're going to get for sine, you'll get zero, you'll get a negative one. So negative negative one is positive one, that one works out. If you try pi halves, sine is one, cosine is zero, so that one works out. And if you try three pi halves, you'll end up with, for sine, of course, you're going to get negative one. For cosine, you get zero, so that doesn't work out here. So the actual solutions are pi and pi halves. This agrees with what we saw earlier, but I just want to show you different ways we could solve this same equation. The Pythagorean equation basically was unavoidable here, but we can't solve it using multiple angles. And as long as you are good about your period here, remember the two pi, you can get all the solutions. And this provides perhaps a simpler approach to the approach we tried earlier. So this brings us to the end of lecture 23. It also brings us to the end of our seventh unit about trigonometric equations in our lecture series. We'll see more trigonometric equations later on and other trigonometric problems. These type of things show up in calculus all the time. So it's very important that students have this skill set developed here from chapter seven. If you learned anything while watching these videos for chapter seven, please hit the like button. Subscribe if you wanna see more videos about trigonometry or other mathematical topics in the future. And as always, if you have any questions, feel free to post them in the comments below and I'll be happy to answer them at my soonest convenience. Bye everyone.