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About this Lesson
 Type: Video Tutorial
 Length: 10:35
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 114 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)
Now that you have learned how to solve simple trigonometric equations, Professor Burger will show you how to solve trig equations that have a coefficient in the argument (e.g. sine of 2X versus just sine of X). These are also called multiple angle equations. In evaluating these trigonometry equations, you will generally only be asked to find solutions in one or two periods of the functions, so you will not have an infinite number of solutions (most often, you'll solve for solutions between 0 and 2*pi). Sample problems from this lesson include (2*sin 3*theta) = 1 and tan^2(2X) = 1 and sin2x*tan2x + sin2x = 0.
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
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Trigonometric Identities
Solving Trigonometric Equations
Solving Trigonometric Equations with Coefficients in the Argument Page [1 of 3]
Let’s take a look at some more exotic trig equations that we want to solve, but now I’m going to put a little extra
restriction on them so we don’t get infinitely many solutions. In these questions let’s just find all solutions that are
between zero and 2? . So remember what that means. What that means is you are allowed to equal zero and you
can go up to 2? but not equal 2? . So basically we’re looking from zero through 60. I just want answers that are inbetween
that range and nowhere else.
Well, the first one I want to take a look at is 2sin3? =1. How would you solve this? Well, the first thing I would do is
try to get the sine thing all alone, so let’s divide everything through by 2, and I see that
sin 3 1
2
? = . Now since I have
that 3 there, what I’m going to do is I’m just going to first of all solve for just 3? , and at the very end of the day I’ll just
divide everything through by 3. So first I want to find out sine of “blank” equals
1
2
; what’s the “blank”? And
remember I’m only between zero and 2? , so if you think of the sine function, what have we got going on here? Well,
I want this to equal
1
2
so between zero and 2? there are only two answers. One is right here and one is right here.
One of them we should know. We should recognize that, in fact, the sine of 30° or
6
?
radians actually has
sin 1
2
= .
So that’s this one. But then how do we get this other one? Well, by symmetry again. You can see that these guys
actually have to be sort of symmetric around here, so what I should do is, since this distance right in here is
6
?
, that
means that this distance right here must be
6
?
. So I’ll take ? and subtract off
6
?
and that will give me this angle. So
one solution is—and I’ve got to be careful here; I’ve got to write this whole thing: 3
6
? =? . But the other answer is
right here, and how do I get that? It’s just this symmetric mirror thing. So I take ? . So 3
6
? =? ?? . Let me say that
again. The idea is that these two points, these two answers, are going to be symmetric. So whatever this distance is
here, this distance must be the same. So if this is
6
?
, this is
6
?
. So how do I find that exact value? Well, I take ?
and subtract the
6
?
. So if I do that, I see
6 5
6 6 6
? ?? = ? . So those aren’t the answers because that’s just what 3?
equals. What does ? equal? We’ll have to divide both sides by 3, and I see
18
?
. And the other answer is
5
18
?
, and
that’s what ? equals. So when you have a multiple angle in here, you just solve for the multiple angle first and then
just divide through and you get the actual value like in this example.
And again, the same rules apply: I’m only looking for answers that are between zero and
2
?
. How about
tan2 2x =1. Well, if tan2 2x =1, that means that tangent alone must equal ±1, taking positive and negative square
roots. So where does that happen? Here’s a rough sketch that I made a while back for the tangent function. Now I’m
just looking from zero to 2? . So where do we see—so the only window I’m looking at here is this. And where do I
see that this thing actually equals 1? Well, it equals 1 it looks like in two places: one over here and one over here.
And where does it equal 1? It equals 1 in two places: one over here and one over here. So I have to find all those
values. There should be four answers to this.
Trigonometric Identities
Solving Trigonometric Equations
Solving Trigonometric Equations with Coefficients in the Argument Page [2 of 3]
Well, first of all, where does it equal 1? Well, that’s when this value is
4
?
or 45°. So in fact what I see here is that
tan 2x =1 when 2x equals either
4
?
, which is 45°, or—how do I get this solution? Well, it’s just the translate of this,
“chunk” over by ? . The period here is ? . So if I add ? to that, I should get this value right here. So this is just
going to be
4
? +? , which equals
5
4
?
. So what does x equal? Well, x equals—if I divide through by 2, I see
8
?
and I
also see
5
8
?
.
Those are the values where tan 2x =1. But what about the 1? Well, to find that, tan 2x = ?1, these are the values
that are sort of right in here. And how would I find that? Well, what I can do is—well, the picture is maybe a little
small so it’s going to be difficult to see, but if you think about it on the angle board, the reference angle is still going to
be 45° or
4
?
but now where am I? Well, where is the tangent negative? All Students Take Calculus. So here and
here. That’s where tangent is negative and the reference angle is
4
?
. So how do I find this angle? Well, this angle is
going to be 180, or ? , minus
4
?
. So one answer is 2
4
x=? ?? . That’s how I get this red angle, by taking 180 or ?
radians and subtracting off that, which is
4
?
. So there are the tangents of 1. And the tangent of the other 1 over
here, how could I get that? I could just take 2? and subtract
4
?
. So the other answer is 2
4
? ?? .
So what do I see? I see that 2x is either equal to—this is
4 3
4 4 4
? ?? = ? , or this answer, which is
8 7
4 4 4
? ?? = ? .
So that’s what 2x equals. So what does x equal? Well, x will equal half of these things. And so what I’m going to
see here is
3
8
?
and
7
8
?
. So those are the two answers that make tan 2x = ?1. But don’t forget the answers that
make tan 2x =1. So in fact, this equation actually has four answers in the range from zero to 2? . And I found them
by factoring carefully and then looking at the graphs and actually finding the solution.
Let’s try one last one. How about this: sin 2xtan 2x+sin 2x=0, and I want to solve that. So I can see that I can
factor out a common factor immediately of sin 2x , so let’s do that. So sin 2x(tan 2x+1)=0 . And so if I have a
product of things equaling zero, either sin 2x = 0 or tan 2x = ?1. Okay, so what are the solutions? Well, in fact, if I
could only reach that thing right down there, I could actually show you. We already did this one, because this one,
tan 2x = ?1, we just solved that. We saw the answers were in fact—we just recorded that here—we’ve already seen
that x would have to be
3
8
?
or
7
8
?
, because tan 2x = ?1 is exactly what we just did on the previous problem. So
we did all that work and I divided through by the 2 and I get that, great. So all we have to do now is this one: where
does sin 2x = 0 ?
Well, where does the sine equal zero? It equals zero at ? and 2? and 3? and ?? and all the multiples of ? , but
I’m only interested remember, in this particular problem, in answers between zero and 2? . So between here and
Trigonometric Identities
Solving Trigonometric Equations
Solving Trigonometric Equations with Coefficients in the Argument Page [3 of 3]
2? , but I’m not allowed to equal 2? , what’s the answer? Well, the answer is just zero and ? . Zero and ? . And so
if the answer is just zero and ? , that means that 2x will equal either zero or ? . So what would x equal? Well, x
would be—if I divide through by the 2, I see zero and
2
?
.
So those are the values that make sin 2x = 0 . These are the values that make tan 2x = ?1. And so what are the
solutions to all of this? Well, zero,
2
?
,
3
8
?
and
7
8
?
. So this equation has four solutions. There are four solutions to
this in the range from zero to 2? . And again, the way I found the solutions was first by factoring and then through
careful analysis of the more simple type trig equations that we talked about earlier.
All right, try these and good luck. These do take a little bit of time and practice, but with a little bit of care you can all
master them—maybe even me! I’ll see you soon.
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