 Welcome back to our lecture series Math 1060, Trigonometry for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Angel Misseldine. In lecture 23, we're going to finish off our unit 7, about solving trigonometric equations. In this lecture, though, 23, we're going to focus on situations where the angle in play might be something like 2 theta, or 3 theta, or 4 theta. Basically, if we change the period, because after all, putting the coefficient from the angle changes the period, 2 theta cuts the period in half, 3 theta cuts it in third, 4 theta changes it by a factor of 4. If we change the period, how does that affect the final answer? And it does have a big effect, because after all, trigonometric solutions repeat themselves over its domain. And so if you change the period, that changes the frequency in which solutions repeat. And so you might have more or less solutions based upon that period change. So special care must be taken when there is a period change to a trigonometric function. So let's look at two examples in this video here, where not a lot's going on other than that period change. Let's find all the solutions for cosine 2 theta equals the square root of 3 over 2, and we're going to solve this in degrees right here. Now ignore the angle for a moment. We're really just trying to think of, when does cosine equal square root of 3 over 2? This would happen in the first quadrant, because it's positive. It would also happen in the fourth quadrant, because again, the ratio is positive. In the first quadrant, cosine is equal to root 3 over 2 at 30 degrees. But that's only in the first quadrant, like in a single rotation, if we go from zero to 360. You'll notice here that there's no constraint on the domain here, so we want all possible solutions. So we could take 360, we could also take 390, because 360 plus 30 is coterminal to 30. So we could add any multiple 360 here. So we're going to slap on a 30 degrees plus 360k. That's if we get the solution from the first quadrant. In the second quadrant, excuse me, in the fourth quadrant, we're looking for the angle that references 30 degrees, which is 360 take away 30. So that would be 330 degrees, all right? In which case, it's not just 330 degrees, it's also like negative 30 degrees. It's anything coterminal to 330 degrees. So we have to throw in a 360k in there as well. So we're looking for the general solution. So far, all of that's great. We never have used the period change whatsoever until this moment. Once we have the general solution ignoring the period change, now we're going to consider the period, because we're trying to solve for theta, not for two theta. So we have to divide the left-hand side by two, so we get a theta. But we have to divide the right-hand side by two as well, for which how division works here, we distribute on the 30 degrees and onto the 360 degrees. So you'll notice that we actually get theta equals, 30 degrees cut in half is 15 degrees. But we also have to take 360k and divide that by two, and we get 180k. And we do that for the second part as well. 330 degrees cut in half is 165 degrees, but then the 360k gets also cut in half. And so we end up with 180k, like so. And so then we still have these representatives of solutions, like 15 degrees represents the general solution, so does 165. But the period has also been changed, right? We have 180 degrees as our period now, right? And that makes sense. If you had two theta, of course, as inside your function, which we do right here, that two, that coefficient of two, which we called B in the past, that would change the period to be two pi over two, which is pi radians, or since we're working with degrees, this would be 180. So this number right here is the new period while 360 was the standard period, the old period. Now, and so we see that solutions will show up more frequently. Consider the graph right here of y equals cosine of two x, where the x coordinates right here and the y axis is right there, right? Because the graph is now 180 degrees periodic, we see that one complete cycle occurs from zero to 180 degrees. So you get one solution right here at 15 degrees, and you get another solution here at 165 degrees. But for the next interval, if we went up to 360, we go from 180 to 360 like so, we end up with a solution right here. We also get a solution right there. So if we had changed the problem, if we had changed the problem, we're like, hey, we don't want just any solutions, we just want the Thetas between zero degrees, but less than 360 degrees. So if that was the stipulation there, how does that change things? Well, you get 15 degrees, you get 165 degrees, but then you get the next one, right? The next one would be 15 degrees plus 180. You also get 165 degrees plus 180 because that second period would show up before 360. And so we have to compensate for those as well. So you get 15 degrees plus 180 degrees, of course, which is 195. And then you get 165 plus 180 degrees, which is 345. Like so, and so this is the thing you have to be very cautious about, that when you change the period, it changes the representatives of the general solution, but it also changes the period of the general solution. So they show up more frequently in this case. You have to compensate for that. Or if you were asked what are the solutions in a certain domain like 360 360, we have to be very careful that we pay attention to the period. That's what one has to be careful about when you change the period here. Let's do another example. Let's find all the solutions to tangent of 3x equals one, and we're gonna do this one in radians. Well, what does tangent equal one? Well, the standard period for tangent, I should mention, is equal to pi, it's not two pi. And so because of this right here, b equals three, we see that the period is gonna change to be pi over three, because you divide that by three. So it repeats itself every pi thirds. So when you, we'll get to that in a second. When you solve tangent equals one, tangent equals one in the first quadrant when sine and cosine are the same thing that happens at 45 degrees or pi force here. You also get in the second, the third quadrant, excuse me, the reference there, but the general solution will be 3x equals pi fourths plus pi k, like so. Because again, for tangent and cotangent, the standard period is pi radians or 180 degrees. For sine, cosine, secant and cosecant, the standard period is 360 degrees or two pi radians. So for tangent, you do need to have that pi k right here. And you don't have to worry about the angle of the third quadrant, which of course in the third quadrant, pi fourths plus pi, that's gonna give you the five pi fourths. That's the angle in the third quadrant that we're thinking of. It really incorporates all of that. So we have to divide by three. So we divide by three, divide by three. And we see that the general solution would be x equals pi twelfths plus pi thirds k. Where you'll notice pi thirds is the change, the modified period for tangent of three x, like so. And so when it comes to this, the general solution is actually not so bad. You just make sure you throw on the period and divide the period by whatever that coefficient changes. But what if, right, this is actually the tricky part. What if they ask us that x needs to be less than two pi but greater than zero. So it's like in one rotation of the circle, what do we get? All right, well in that situation, we're definitely gonna get pi twelfths, right? But we would also get pi twelfths plus a multiple of pi thirds. We'd also get pi twelfths plus two pi over three. So two multiples of that. Because after all, pi thirds happens, that's gonna get us up to pi, right? We need to keep on going. So the thing is, since we cut the period in thirds, it's gonna repeat itself three times as we complete one period. But again, we wanna go up to two pi. We also would need pi twelfths plus pi, three pi thirds. We need pi twelfths plus four pi thirds. We'd also need pi twelfths plus five pi thirds. And you could add those facts together to simplify them. But the point is, as you go from zero to pi, there would be six solutions in all as opposed to the standard tangent, which would have two solutions from zero to two pi. Where did six come from? Well, we take the two solutions from the standard tangent and then we times that by three because we cut the period in by third, thus raising the frequency by a factor of three. So finding the general solution when you change a period is actually pretty nice because you have to just change the period in the general solution. But if it wants to enumerate all the answers from zero to two pi, you have to be meticulous to make sure you get each and every one of those. And really you should simplify the fractions here. You should take pi twelfths plus four pi thirds and write as a single fraction. I'm not doing that here for the sake of time. But be aware of that. That's something you might be expected to do.