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PreCalculus: Using DoubleAngle Identities 
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Int Algebra: Using the Quadratic Formula
About this Lesson
 Type: Video Tutorial
 Length: 14:55
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 161 MB
 Posted: 01/22/2009
This lesson is part of the following series:
Trigonometry: Full Course (152 lessons, $148.50)
PreCalculus Review (31 lessons, $61.38)
Trigonometry: Trigonometric Identities (23 lessons, $26.73)
Trigonometry: Solving Trigonometric Equations (5 lessons, $7.92)
Pre Cal: Solving Trigonometric Equations (5 lessons, $7.92)
Sometimes, trigonometric equations cannot be factored. To solve these equations, Professor Burger shows you how to apply the quadratic formula to find solutions to these equations. This is a multistep process that starts with simplifying the equation. After the equation is simplified, you will be able to solve the quadratic formula and then enter that answer in to solve for X. In the lesson example, Professor Burger uses both a calculator and graphing to ensure he has the correct points. This lesson explains the covered material by walking through sample equations 3sin^2(2x) + sin (2x)1 = 0. This equation has 2 solutions over the interval from zero to 2*pi. This lesson is loaded with warnings about easy mistakes to make and pitfalls to be wary of when evaluating problems like this.
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Precalculus. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/precalculus. The full course covers angles in degrees and radians, trigonometric functions, trigonometric expressions, trigonometric equations, vectors, complex numbers, and more.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
About this Author
 Thinkwell
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11/13/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
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Solving Trigonometric Equations Using the Quadratic Formula
Sometimes when you have an equation that you want to solve that has trigonometric function, and you try to factor it because it looks like it should be factorable somehow, sometimes you just may not be able to factor it. It’s sort of a sad truth of life sometimes. But in that case we can use the quadratic formula or some other trick. Let me actually show you an example where the thing looks so factorable and yet it turns out we’ve got to do something else.
So, how about this exotic equation. Suppose we have 23sin2sin210xx+?= and now I want to solve that for x. Okay, now what’s the first thing I notice? Well, I see 2sin2x and a sin2x, so it sounds like, gee, maybe I could sort of factor this thing. It looks like a quadratic, and maybe I could sort of write it out somehow. Now, you may be more comfortable thinking about it this way—and in fact, this might not be a bad idea. In fact, let me just write it out. I could write this. I could replace the 2sin2x by just something—I’ll call it s for something. So I could write this as 3 times something squared, plus just the something, minus 1 equals zero, where we should remember what the something is. The something is sin2x. This sometimes makes it easier to sort of see what’s going on in terms of what you have to factor without worrying about all those pesky trig functions until the very end when you really need to factor completely.
Well, now you might want to try to factor this, and try to put two binomials together and make combos and stuff like that. It turns out you won’t be able to succeed because this actually can’t be factored in some nice way. So what do you do? Do you throw your hands up? No! You use the quadratic formula to let you solve for x. So if we do that, let’s use the quadratic formula right now. So what does s equal? Well, s equals negative b—so again, the roles here, this would be playing the role of a; b would be the invisible +1; and c is the notsoinvisible 1. So I see s equals negative b—so that’s –1, 2b±, which is just 1 squared, which is 1, 4ac. Well, that’s 4 times –3, which would be 12, but that negative sign with a 2 makes this a +12, and that’s all under the square root, all divided by 2a, which would be 6. So I see that s actually equals two numbers. It equals 1316?+ and 1316??.
Now, let me actually evaluate those numbers numerically—at least approximate them numerically because we’re going to be crunching them with the two signs. So I see that s either equals—when I put the plus sign in there, 1316?+, if you computer that on the calculator you see 0.43425 and it keeps going. And if you put in the minus there, 1316??, that will be a negative number, and it turns out that the negative number is 0.7675918 and it keeps going forever, too. So now I’ve solved the s part of the problem. Unfortunately, that wasn’t the question. The question wasn’t “find s.” The question was, ”find x.”
So I have to remember now what x equals. So now I’m going to see two here two separate, little teeny simple trig equations that I have to consider. So let’s now consider those trig equations. So what I see now is sin2xequals that; sin2xequals that. So let’s write those out here. In fact, let’s just solve them separately. But let me just write them both out before we forget what they are. So we have two things to do. We have sin2xequals, and the first one is 0.43425, so we have to solve that. But then we also have to solve sin2xequals the other solution, which is 0.76759 and such. So let’s solve this one first. Let’s take a look at this one here first.
Well, how do I solve that? By the way, just like in these other problems we’ve been looking at, we want to just look for the solutions that are between zero and 2?. So in fact what I want here are the solutions to this original equation, where the x is going to live inside of zero to 2?, so anywhere from zero to 360°, if you wish. So I only want to find those x’s that are in there.
Okay, let’s see. I want sine of something to be 0.43something. How many solutions should there be? Well, here’s the sine curve. Let’s just take a look at that and visualize it. I’m only looking from zero to 2? so I don’t look at those things. I just look at that complete cycle. And I want to know, where does that have a height of 0.4something? Well, 0.4something is around here, and I see there are two places. Someplace right over here, and then someplace that’s
? minus that. So I’ve got to first of all find that reference angle, and I can just do that on a calculator by using the arc sine or the inverse sine function. So if you do that, you take your calculator and use inverse sine, and what I see is that 2x will equal—well, how many radians? Well, it turns out it’s 0.44921 and so on radians. That’s how many radians this whole angle is. But remember I want actually half of that, so we just have to divide through by 2. So if I divide through by 2, what do I see? Well, I’m going to see that divided by two, which is 0.224607 radians. There’s one answer. All that work and we just have one answer. Boy, you’d think after all that work you’d have a ton of answers.
Okay, there’s another solution to this thing, right? Remember what we just saw was where we equal 0.43something, which is over here. But there’s another solution. That other solution is way over here. So how do I find that solution? Well, if this angle right here gives me a value for sine of that, then the angle that’s sort of symmetric but a little bit in toward the ? radian by the same exact amount will give me another value by symmetry. So all I have to do now is take ? and subtract off the angle that I found here. So I have to take ? and subtract that number, so the other solution is 2x?=minus that number that we just found. Well, what is that? Well, I take ? and then you subtract off 0.44921, and that equals 2.6927. So to find the other solution I have to divide this by 2, so I see x equals 1.3461 radians. So there’s a second answer. And these are the two answers to this particular equation.
But now I’ve got to do the exact same thing with this one! Oh, boy, it never ends. It never, ever ends. So how do I do that? These two solutions that we just found were for this red possibility, but now we also have to consider this possibility, so that’s another probably two solutions; that would be my guess. Let’s take a look at the sine curve again and see where those things are.
So I’m only looking between zero and 2? and I want to know, well, for which angles do I get 0.76? So 0.76 is around here somewhere, and you can see there are actually going to be two answers again. One is going to be over here and one is going to be over here. So let’s see if we can find those. How the heck are we going to find those values? Well, we’ve got to be a little bit careful here. So I could use the arc sine function again. What I do is I put in the inverse sine of that, and what does that give me? That has a negative sign in front of it. Oh, that negative sign is so critical, so critical. The whole would change. You know the whole world would be a completely different place if I wouldn’t put a negative sign right there? It’s true.
Okay, I get the answer of 0.87507something. Now, what does that mean exactly? Let’s think about that, because that seems to be a negative answer, so that’s a little disturbing. Why is that a negative answer? Because the answer should be sort of living around here. Well, we have to remember what the inverse sine does. The inverse sine operates on a slightly different region than we’re looking at. We have to now consider the following. The inverse sine is operating right along here. This is where the inverse sine is operating, and this is 2?, and this is 2?. So the answer that we found for what angle gives me 0.76something is actually this negative angle right here. Well, that’s not exactly the one we want. We know which one we want. We want it between zero and 2?. So what should I do?
Well, let’s think about it for a second. We have the sine function here, and I’ve got this piece sort of way down here, not exactly where I wanted it, but what can I do? Well, notice if I were to add 2? to that answer, where would that put me? Well, that would put me right over here. So if I add 2?, that shifts me right over to here, and now that’s one of the answers I want, so that looks good. So what I have to do is take that answer I just found, which I know is wrong, but I should add 2? to it. That will shift me over to the right region I’m looking at. That will give me this wing here. Why that wing? Because notice that that’s sort of living on this part of the wing, so when I shift it I’m going to live on the corresponding part of the wing. I won’t live over here; I’ll live over here. So let’s add 2? to that, and what I get is that 2x equal 5.40811something. And notice that’s in the right range because 5.4something is what? It’s going to be between ? and 2?. ? is 3.14something, 2? is 6pointsomething, so this is right somewhere inbetween it, so that looks good, at least visually. And so therefore if I want to find out what x equals, I just divide that by 2 and I see that one solution for x here is going to be 2.70405 and it keeps going for a while. Okay, so there’s one solution.
But don’t forget, in this one just like the other, there must be two solutions. What we just found was this solution right here. How would I find this solution? Well, it’s going to be a little tricky because I have to figure out what this angle should be, and all I know is that these things are going to be symmetric, so they’re right around here. So I know this value; how do I find this value? Well, let’s think about it. If I know this value, I know its distance away from 2?. So what is its distance away from 2?? Well, that distance actually is exactly the distance that we already found. It turns
out it’s going to be this value right here. So if you take 2? and subtract this angle, you take 2?, and now we’re going to undo a little what we just did. Take 2? and subtract off the angle we found here, 5.40811 and so forth. What we get is this thing without the negative sign, so we get 0.87something. That tells me this little difference here, and that’s how much I have to add to this side to bring it in.
Let me say that again. I’ve got this angle right here. I just found that. But I want to find this little teeny piece there because that’s how much I have to add to ?. So I took 2? minus that angle, found that value to be 0.87507 and so forth, and now I just take ? and add that amount and that should be my other answer. So I just add this to ?, plus ?. Isn’t this amazingly involved? And so what that tells me is that the other answer is 2x equals 4.0166something. And notice that also makes sense because that’s just a little bit bigger than ? but not as big as the 5 answer we got before, so that looks good. I just divide by 2 now, finally, to get x alone. Phew! This is exhausting, isn’t it? But yet if you’re careful, you can nail everything. You see 2.0083something radians. These are all in radians, of course.
So these are the two solutions from the interval zero to 2? for this equation. We already found the two solutions from zero to 2? for this equation. And we saw that those two equations together, those solutions, actually give us the solutions with the original quadratic. So I used the quadratic formula. I found the two values for s, but was actually sine. I had to solve both those sine equations separately, got two answers for this and got two answers for that. There are four solutions to this baby. But really take it careful and easy when you have quadratics and trig things because, as you can see, there are lot of different cases that you have to look at very, very carefully, including the graphs, to make sure you get those right points. And also caution when you’re taking the inverse sine function. Remember the region where the inverse sine function is defined and make sure that you’re finding the right spot.
Good luck. You can do it.
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