WEBVTT 00:00:07.103 --> 00:00:15.197 Today we're going to look at trigonometric equations. A trigonometric equation has a trig function as part 00:00:15.199 --> 00:00:24.039 of the equation. There are two types: an identity which we studied previously and conditional equations 00:00:24.041 --> 00:00:34.880 which we're going to study today. Let's review a minute. What is an identity? An identity is true for all 00:00:34.882 --> 00:00:41.630 solutions of the variable. Now I'm going to write two examples on the board, and I want you to be 00:00:41.632 --> 00:00:49.127 copying these because we're going to need them in some other examples. So the first one that you should 00:00:49.129 --> 00:01:00.131 recognize is the sine of theta squared plus the cosine of theta squared equals 1. This is our famous 00:01:00.133 --> 00:01:07.995 Pythagorean identity, and we will use it. The second one that I'm going to write here, be copying 00:01:07.997 --> 00:01:23.497 remember, is the cosine of 2 theta equals 2 cosine theta squared minus 1. So what is a conditional 00:01:23.499 --> 00:01:32.493 trigonometric equation? Where identities are true for all solutions of the variable, a conditional trigonometric 00:01:32.495 --> 00:01:41.997 equation is only true for specific values of the variable. The example that we're going to start with today is 00:01:41.999 --> 00:01:55.996 this one. 2 cosine theta plus 1 equals 0. We will get started in just a minute. 00:01:58.290 --> 00:02:06.792 We're going to look at trigonometric equations that are linear in form. The example that we spoke about 00:02:06.794 --> 00:02:16.544 earlier is 2 cosine theta plus 1 equals 0, and we want to do two things from our directions. We want to 00:02:16.546 --> 00:02:25.546 determine a general formula for the solutions of the equation. We're going to let k represent any integer. 00:02:25.548 --> 00:02:35.754 Then as a b part, we're going to determine the specific solutions, if any, on the interval 0 to 2 pi. 00:02:35.756 --> 00:02:42.094 So in looking at this equation, the first thing you might ask yourself is, "What do I do with it? I don't know 00:02:42.096 --> 00:02:52.936 where to begin to solve this equation." So you might come here on the side and use substitution. You could 00:02:52.938 --> 00:03:04.439 use any letter, u or x or y or w, and let that be the trig function. Making the substitution, you should see 00:03:04.441 --> 00:03:13.188 an equation that you recognize from algebra. What do we do to solve this equation? Well, very simply, 00:03:13.190 --> 00:03:23.805 we subtract the 1, and then we divide by 2 on both sides of the equation. So in algebra, we would have this 00:03:23.807 --> 00:03:32.553 solution. So now what do we do with our trigonometric equation? The exact same thing. We're going to 00:03:32.555 --> 00:03:48.078 subtract 1 and then divide by 2. Ok, unlike the algebra equation, where we were finished, we now have 00:03:48.080 --> 00:04:01.076 to solve for the variable theta. We now want to find the general formula for all of the solutions. Well, in 00:04:01.078 --> 00:04:09.828 order to do that, we need to do a couple of things. The first thing we need to do is find the reference angle, 00:04:09.830 --> 00:04:22.330 theta r, so I'm going to locate. Let's remember our right triangles, and let's remember that the cosine function 00:04:22.332 --> 00:04:34.827 is the ratio of the adjacent side, so we're going to let 1 be the adjacent side, 2, ratio of the adjacent side 00:04:34.829 --> 00:04:46.573 over the hypotenuse. So we have 1 divided by 2. That means the opposite side is the square root of 3. 00:04:46.575 --> 00:05:02.076 Ok, what is that reference angle? Well, it is pi over 3. Now in this problem our ratio is negative, so pi 00:05:02.078 --> 00:05:12.745 over 3 is not one of our answers. [In] which two quadrants is the cosine function negative? Quadrant two, 00:05:12.747 --> 00:05:23.496 which will give us one solution, and quadrant three, which will give us a second solution. Ok, reference 00:05:23.498 --> 00:05:39.246 angle pi over 3. Now let's see what the general formula is. The angle one, quadrant two, we will find by 00:05:39.248 --> 00:05:58.454 subtracting the reference angle from pi. So we have pi minus pi over 3, which is 2 pi over 3, plus the 00:05:58.456 --> 00:06:11.206 period of our cosine function is 2 pi, and we will multiply that times k. This will give us a set of solutions. 00:06:11.208 --> 00:06:27.571 What about the second angle, theta two? Well, in quadrant three, theta two will be pi plus the reference angle, 00:06:27.573 --> 00:06:48.076 so that'll be pi plus pi over 3, which is 4 pi over 3, plus again 2 pi times k. That is the answer to the a 00:06:48.078 --> 00:06:58.329 part of our problem. What do we do for the b part? We want to find specific solutions on the interval 00:06:58.331 --> 00:07:11.101 from 0 to 2 pi. All we have to do, we're going to begin by letting k be 0. 0 times 2 pi of course is 0. 00:07:11.103 --> 00:07:26.102 That's going to give us a solution of 2 pi over 3. Same thing, let k be 0 here. That gives us 4 pi over 3. 00:07:26.104 --> 00:07:35.877 Ok, so do we let k be 1? Well, let's see what happens. If k is 1, then we would be adding 2 pi to each, 00:07:35.879 --> 00:07:49.124 2 pi over 3 and 4 pi over 3. Both of those answers are not in our interval, 0 to 2 pi. So we see that on that 00:07:49.126 --> 00:07:53.378 given interval we have two solutions. 00:07:55.560 --> 00:08:03.312 Let's look at this example. We want to determine a general formula for the solutions of the equation 00:08:03.314 --> 00:08:13.820 tangent theta plus 1 equals 0. In our formula we're going to let k represent any integer, and then in the 00:08:13.822 --> 00:08:22.247 b part, we're going to determine the specific solutions, if any, on the interval 0 to 2 pi. Now, this example 00:08:22.249 --> 00:08:32.017 is an example of a trigonometric equation that is linear in form. When we first look at it you may not 00:08:32.019 --> 00:08:44.862 be sure exactly what to do with it. So you can very easily come here to the side, let u be your trig function, 00:08:44.864 --> 00:08:57.954 tangent theta. Making the substitution, we have the equation u plus 1 equals 0 and solving we subtract 1 from 00:08:57.956 --> 00:09:07.384 both sides. We see that u is negative 1. We're going to do the exact same procedure for our trig equation. 00:09:07.386 --> 00:09:20.226 We're going to subtract 1, and then we want to find all the solutions of theta. The ratio is negative, 00:09:20.228 --> 00:09:27.729 but to work this problem, we're going to first find the reference angle. So let's come over here to the side 00:09:27.731 --> 00:09:46.252 and find theta r. In our right triangle, we're going to set up tangent is the ratio. Let's rewrite that here. 00:09:46.252 --> 00:09:56.435 Tangent of theta r equals positive 1. We want to know where the tangent is positive. So tangent is the 00:09:56.437 --> 00:10:08.530 ratio of the opposite side to the adjacent side. Those sides are each 1. By the Pythagorean theorem, the 00:10:08.532 --> 00:10:19.464 hypotenuse is now the square root of 2, and we recognize the reference angle to be pi over 4. 00:10:19.466 --> 00:10:31.798 Back to our problem. Tangent theta is negative 1. Ok, where is the tangent of theta negative? There are 00:10:31.800 --> 00:10:43.393 two quadrants in which the tangent function is negative. Quadrant two, we'll call it theta one. Quadrant 00:10:43.395 --> 00:10:57.572 four, we'll find that solution and call it theta two. To find the first angle in quadrant two, we need to 00:10:57.574 --> 00:11:16.328 subtract the reference angle from pi, so pi minus pi over 4 is 3 pi over 4, and then quadrant four, we want to 00:11:16.330 --> 00:11:36.786 subtract the reference angle from 2 pi. 2 pi minus pi over 4 is 7 pi over 4. So we come back here, and 00:11:36.788 --> 00:11:50.377 we want to write our general formula. We have 3 pi over 4 plus pi k because the period of the tangent 00:11:50.379 --> 00:12:05.877 function is pi, and 7 pi over 4 plus pi k. Ok, but looking at this, this doesn't seem very concise, and it's not 00:12:05.879 --> 00:12:20.245 because if we let k be 1 and add pi to 3 pi over 4 we end up with 7 pi over 4. So the better way to write the 00:12:20.247 --> 00:12:38.428 general formula is simply 3 pi over 4 plus pi k. That will be our answer for the a part, the general formula. 00:12:38.430 --> 00:12:56.698 Now in the b part, we will begin by letting k be 0. That of course gives us 3 pi over 4. Let k be 1. As we 00:12:56.700 --> 00:13:13.701 previously said, pi plus 3 pi over 4 is 7 pi over 4. But can we let k be 2? I don't think so. 2 pi plus 3 pi over 4 00:13:13.703 --> 00:13:28.697 is outside of our interval. We have 2 solutions, 3 pi over 4, 7 pi over 4. What you want to remember about 00:13:28.699 --> 00:13:36.851 this problem is that when you're solving with the tangent function you're going to work it very differently 00:13:36.853 --> 00:13:44.300 than if your trig equation has the sine or the cosine function. 00:13:46.900 --> 00:13:56.942 We're going to look at this example. The sine of 2 theta equals 1 over the square root of 2. Our directions 00:13:56.944 --> 00:14:07.083 tell us to find the specific solutions, if any, on the interval from 0 to 2 pi. Now, this trig equation is linear 00:14:07.085 --> 00:14:15.927 in form, but it's a little different than some of the others that you may encounter in your homework, and 00:14:15.929 --> 00:14:27.268 that's because there is a constant multiplied times our angle theta. So what we want to think about is what 00:14:27.270 --> 00:14:38.696 solutions of this angle, what solutions when we take the sine of that angle, will equal 1 over the square 00:14:38.698 --> 00:14:48.791 root of 2. Back here, at the side, we can find the reference angle, but you may know it immediately because 00:14:48.793 --> 00:14:57.881 1 over square root of 2 sine function is very ... It's in our family, and it's very obvious to us. Sine function, 00:14:57.883 --> 00:15:14.633 ratio of the opposite side to the hypotenuse, we see the reference angle is pi over 4. I am going to set this up 00:15:14.635 --> 00:15:24.132 using a general formula to begin with, even though the problem does not call for me to give that in my 00:15:24.134 --> 00:15:38.089 final answer. So we'll write 2 theta equals pi over 4 plus 2 pi k. Now that's only one quadrant in which the 00:15:38.091 --> 00:15:48.183 sine function is positive. Ok, let's think about this. [In] which other quadrant is the sine function positive? 00:15:48.185 --> 00:16:00.027 Quadrant two, correct. That will give us a second solution which we will find by subtracting the... oops 00:16:00.029 --> 00:16:24.276 watch the mistake here. Theta two is pi minus the reference angle, so pi minus pi over 4 which is 3 pi over 4. 00:16:24.278 --> 00:16:34.775 Now if we did not have this 2, we would be done as far as finding a general formula, but we have to deal 00:16:34.777 --> 00:16:45.324 with the 2. The 2 is multiplied times theta. We need to take it to the other side, so we're going to divide. 00:16:45.326 --> 00:16:58.914 It's very important that we divide all four terms by the 2. Pi over 4 divided by 2 is pi over 8. 2 pi k divided 00:16:58.916 --> 00:17:18.914 by 2 is pi k. 3 pi over 4 divided by 2 is 3 pi over 8, and again, 2 pi k divided by 2 is pi k. Now we can start 00:17:18.916 --> 00:17:35.667 finding our specific solutions. We let k be 0. That gives us pi over 8, then 3 pi over 8. Then we'll let k be 1, 00:17:35.669 --> 00:17:53.663 so we're adding pi over 8 plus pi which is 9 pi over 8. Let k again be 1. Add pi plus 3 pi over 8. That gives 00:17:53.665 --> 00:18:04.152 us 11 pi over 8, and of course we cannot let k be 2 because we'd be adding 2 pi to each of the angles 00:18:04.154 --> 00:18:15.150 which is outside of our interval. So we see in this example that we have 4 solutions. 00:18:16.583 --> 00:18:26.675 We're going to look at a trigonometric equation that is quadratic in form and attempt to solve it. Our 00:18:26.677 --> 00:18:36.767 directions tell us to find the specific solutions, if any, on the interval from 0 to 2 pi, and our example reads 00:18:36.769 --> 00:18:49.358 3 times the cosecant of theta squared minus 4 equals 0. I'm not sure how to begin the problem, so I'm 00:18:49.360 --> 00:19:01.859 going to come here to the side. Let u equal my trig function, cosecant theta. Making the substitution, I have 00:19:01.861 --> 00:19:19.474 3 u squared minus 4 equals 0. Then I recall from algebra, I would add the 4, divide by 3, and then one 00:19:19.476 --> 00:19:28.568 more step. To undo the square, I would take the square root on both sides, but I need positive and 00:19:28.570 --> 00:19:43.563 negative square root of 4/3, and of course this simplifies to plus or minus 2 divided by the square root of 3. 00:19:43.565 --> 00:20:06.027 I'm going to do the same thing over here. I'm going to add the 4, divide by 3, take the square root, and I 00:20:06.029 --> 00:20:24.460 end up with the cosecant of theta is plus or minus 2 divided by the square root of 3. I need to solve for theta. 00:20:24.462 --> 00:20:36.890 I'm going to find the reference angle. Here to the side, I'm going to find the reference angle such that the 00:20:36.892 --> 00:20:54.574 cosecant of the reference angle equals the positive ratio 2 divided by square root of 3. In quadrant one, my 00:20:54.576 --> 00:21:05.822 reference angle, I'm going to place the 2 and the square root of 3 at the appropriate sides. I know cosecant 00:21:05.824 --> 00:21:18.073 is the ratio, cosecant of theta is the ratio of the hypotenuse, so that's the 2, divided by the opposite side, 00:21:18.075 --> 00:21:32.194 which is square root of 3. That makes this side 1, and I know that my reference angle is pi over 3. Ok, 00:21:32.196 --> 00:21:41.536 that'll give me one solution. Let's write that solution down for positive 2 over square root of 3 in quadrant 00:21:41.538 --> 00:21:55.466 one, but consider this. Cosecant is positive in quadrant one. It's positive in quadrant two. It's negative in 00:21:55.468 --> 00:22:08.652 quadrant three and negative in quadrant four. So we have four solutions. Ok, pi over 3 is our solution 00:22:08.654 --> 00:22:20.900 in quadrant one. Let's call it theta one. Theta two, our second solution, we will find by subtracting the 00:22:20.902 --> 00:22:37.177 reference angle from pi which gives us 2 pi over 3, our second solution. Third solution, in quadrant three, 00:22:37.179 --> 00:22:57.951 we will find by adding pi to the reference angle. Pi plus pi over 3 is 4 pi over 3, third solution. Finally 00:22:57.953 --> 00:23:13.950 in quadrant four, we will get our solution by subtracting the reference angle from 2 pi, 00:23:13.952 --> 00:23:28.477 and that gives us 5 pi over 3. This trigonometric equation that is quadratic in form 00:23:28.479 --> 00:23:36.224 has four specific specific solutions on the interval from 0 to 2 pi. 00:23:38.069 --> 00:23:46.564 We'll take a look at this example. We want to determine the specific solutions, if any, on the interval from 00:23:46.566 --> 00:23:56.809 0 to 2 pi of the trigonometric equation that is quadratic in form, and our example reads 2 times the sine of 00:23:56.811 --> 00:24:10.311 theta squared plus the sine of theta equals 0. I'm going to make my substitution and let u be the sine of 00:24:10.313 --> 00:24:23.055 theta, so I can readily recognize how to solve this equation. In algebra, substituting u here, if u is the sine 00:24:23.057 --> 00:24:35.673 of theta, then sine of theta squared, will be u squared plus u equals 0. Then I recall from algebra when 00:24:35.675 --> 00:24:47.175 I have this on the left side of the equation I can factor the common factor, u, and after doing that, u times 00:24:47.177 --> 00:25:05.290 2u plus 1 equals 0. The two factors give me 0, so I can set each factor equal to 0 and solve independently. 00:25:05.292 --> 00:25:17.793 First solution is u equals 0. Second solution is u equals negative 1/2. I subtracted 1 then divided by 2. 00:25:17.793 --> 00:25:30.040 So that's what we want to do with this trig equation. The common factor is the sine of theta. Write that, 00:25:30.042 --> 00:25:45.004 then divide. Sine of theta into 2 sine squared is 2 sine theta plus sine of theta into sine theta is 1. Or you 00:25:45.006 --> 00:25:57.001 can simply look at what you have here and just substitute back. For u equals 0 we have sine theta equals 0, 00:25:57.003 --> 00:26:16.251 like this factor, and then 2 sine theta plus 1 equals 0. Solving this one, sine theta equals negative 1/2. 00:26:16.253 --> 00:26:29.957 We have now two trig equations that are linear in form to solve and get solutions for. Sine of theta equals 0. 00:26:29.959 --> 00:26:39.049 Now that's not going to produce, if we are looking for a reference angle, it's not going to produce pi over 3, 00:26:39.051 --> 00:26:49.392 pi over 4, or pi over 6, our special angles because if the ratio for the sine function is the opposite over the 00:26:49.394 --> 00:26:58.235 hypotenuse that will not form a triangle. So what I need to look at is the graph of my sine function. 00:26:58.237 --> 00:27:14.915 Let's graph from 0 to 2 pi, just a quick sketch. Then we will look and see where is the sine function 0? 00:27:14.917 --> 00:27:29.599 Sine function is 0, looking at the x-axis or the x-intercepts, is 0 at 0, pi, 2 pi, 3 pi, 4 pi, and continues, 00:27:29.601 --> 00:27:42.531 but we only want the solutions from 0 to 2 pi. Notice we do not want to include 2 pi, so for this equation, 00:27:42.533 --> 00:27:57.629 sine theta equals 0, we have two solutions at 0 and at pi. Ok, this equation now, we have the ratio 00:27:57.631 --> 00:28:06.626 negative 1/2, so we'll come to the side and find our reference angle. We're looking for pi over 3, pi over 4, or 00:28:06.628 --> 00:28:24.874 pi over 6. So we want to solve the equation the sine of theta r equals positive 1/2. Sine function, opposite 1, 00:28:24.876 --> 00:28:40.174 hypotenuse 2, adjacent side square root of 3, so the reference angle, pi over 6. That's right, but back to 00:28:40.176 --> 00:28:48.672 our problem. This ratio is negative, so we don't want an answer of pi over 6. We want the two quadrants 00:28:48.674 --> 00:29:01.357 where the sine function is negative. Quadrant three, that will give us a solution. Quadrant four, that will 00:29:01.359 --> 00:29:15.540 give us another solution. Quadrant three, we need pi plus the reference angle, pi plus pi over 6, which will 00:29:15.542 --> 00:29:28.129 give us 7 pi over 6, and then in the fourth quadrant, an answer for theta will be 2 pi minus the reference 00:29:28.131 --> 00:29:50.881 angle, which is 2 pi minus pi over 6, 11 pi over 6. Ok, this particular equation then has four solutions. 00:29:50.883 --> 00:30:06.339 Sometimes in writing your solutions you may be asked to put the solutions in ascending order, so just 00:30:06.341 --> 00:30:27.592 double check: 0, pi, then 7 pi over 6, then 11 pi over 6, and that is our final answer for this particular example. 00:30:30.000 --> 00:30:38.761 We're going to look at an example of a trigonometric equation that we will solve using identities. Our 00:30:38.763 --> 00:30:48.512 directions tell us to determine the specific solutions, if any, on the interval from 0 to 2 pi, and our example 00:30:48.514 --> 00:31:00.260 reads negative 3 times the sine of theta plus 3 equals 2 times the cosine of theta squared. The first thing 00:31:00.262 --> 00:31:10.260 we notice about this example is that we have the sine of theta and cosine of theta, two different trig functions. 00:31:10.262 --> 00:31:23.762 So what we need to do is find an identity that we can change either sines into cosines or cosines into sines. 00:31:23.764 --> 00:31:36.696 Let me think. Let me think. We have an identity that says sine squared theta plus cosine squared theta 00:31:36.698 --> 00:31:48.196 equals 1. Oh, that's a good one to use. We have the cosine of theta squared, so let's solve for the cosine 00:31:48.198 --> 00:32:02.946 of theta squared by subtracting the sine of theta squared from both sides. Then we can take this 1 minus 00:32:02.948 --> 00:32:20.150 sine of theta squared, and we can substitute it for the cosine of theta squared. I am going to put parentheses 00:32:20.152 --> 00:32:33.495 here to enter 1 minus the sine of theta squared because I'll need to come back and distribute the 2. 00:32:33.497 --> 00:32:53.271 Let's do that. Negative 3 sine theta plus 3 [equals] 2 times 1, 2 times negative sine of theta squared. 00:32:53.273 --> 00:33:02.519 Ok, let's see what we're going to do now. We need all of the terms on one side of the equation set equal 00:33:02.521 --> 00:33:13.203 to 0 so we can factor. The squared term, sine of theta squared, has a negative 2 in front, so I'm going to 00:33:13.205 --> 00:33:28.454 take that term to the left side of the equation by adding 2 sine of theta squared. Copy negative 3 sine of 00:33:28.456 --> 00:33:44.318 theta, and then subtract 2 from both sides. 3 minus 2 is plus 1. So we have a trigonometric equation that 00:33:44.320 --> 00:34:01.071 is in quadratic form, and we can factor. I notice the factors of 2 sine theta squared will be 2 sine theta and 00:34:01.073 --> 00:34:17.188 sine theta. Factors of 1 are 1 and 1, and then we need minus in both of those factors. Checking, this is 00:34:17.190 --> 00:34:28.026 negative sine theta. This is negative 2 sine theta, which is negative 3 sine theta. Now if you had to, you 00:34:28.028 --> 00:34:44.712 could come to the side and make a substitution for sine theta, and you would get the same factorization, 00:34:44.714 --> 00:34:54.963 but it might make it a little easier for you. Once we have our two factors set equal to 0, all we have to do, 00:34:54.965 --> 00:35:13.711 let's erase this, is solve each factor separately. 2 sine theta minus 1 equals 0. Set that factor equal to 0. 00:35:13.713 --> 00:35:34.962 Then set this factor, sine theta minus 1, equal to 0. Sine theta will then equal, add 1, divide by 2. Here, add 1. 00:35:34.964 --> 00:35:47.461 Ok, sine theta equals 1/2. We'll come to the side, find the reference angle, and then find the two quadrants 00:35:47.463 --> 00:36:03.537 in which the sine function is positive. Sine of the reference angle equals positive 1/2. Quadrant one, 00:36:03.539 --> 00:36:15.288 opposite side over hypotenuse. The adjacent side, then, is square root of 3, and the reference angle is 00:36:15.290 --> 00:36:29.537 pi over 6. Ok, that will give us one solution because this ratio is positive. Sine function is positive 00:36:29.539 --> 00:36:36.996 in quadrant one, same thing as the reference angle. In which other quadrant is the sine function positive? 00:36:36.998 --> 00:36:49.589 Well, that's quadrant two. That will give us a second solution. We find that angle by subtracting the 00:36:49.591 --> 00:37:11.587 reference angle from pi. Pi minus pi over 6 is 5 pi over 6. These are two solutions, solves this equation 00:37:11.589 --> 00:37:23.085 on this interval. Now we'll go to this part. Sine theta equals positive 1. We don't have a reference angle of 00:37:23.087 --> 00:37:37.587 pi over 3, pi over 4, pi over 6 for the sine of theta equals 1. What we need there is the graph of our sine 00:37:37.589 --> 00:38:00.337 function. So I'm going to graph from 0 to 2 pi. Well, let's do this a little better. There we go. That looks better. 00:38:00.339 --> 00:38:17.346 So, our range from negative 1 to 1, we notice that the sine of theta equals positive 1 at pi over 2. So 00:38:17.348 --> 00:38:27.437 summarizing, and we're going to write our final solution. On the interval from 0 to 2 pi, we have 3 answers, 00:38:27.439 --> 00:38:36.687 pi over 6, 5 pi over 6, pi over 2. I'm going to write the three solutions in ascending order, so I'm going to 00:38:36.689 --> 00:38:53.688 write pi over 6, pi over 2, and 5 pi over 6. Again, to solve this equation using identities, we made a 00:38:53.690 --> 00:39:00.900 substitution for cosine theta from our Pythagorean identity. 00:39:03.238 --> 00:39:13.437 This example involves us solving this trigonometric equation, cosine of 2 theta minus 3 equals negative 7 00:39:13.439 --> 00:39:23.686 times the cosine of theta. Our directions tell us to determine the specific solutions, if any, on the interval 00:39:23.688 --> 00:39:36.633 from 0 to 2 pi. We're going to use an identity because this trig function, cosine 2 theta, is different than this 00:39:36.635 --> 00:39:49.134 one which is the cosine of just theta. I want to use the identity for the cosine of 2 theta. There are actually 00:39:49.136 --> 00:40:03.385 three versions to this identity. We have cosine squared minus sine squared. We have 1 minus 2 sine of theta 00:40:03.387 --> 00:40:16.633 squared, and we have 2 cosine of theta squared minus 1. So I want to choose the one that will work the best 00:40:16.635 --> 00:40:27.632 for this problem. Since the equation has a cosine of theta function, we will use this version. I'm going to 00:40:27.634 --> 00:40:44.133 substitute 2 cosine of theta squared minus 1 for the cosine of 2 theta. Let's do that. 2 cosine of theta 00:40:44.135 --> 00:41:00.883 squared minus 1 minus 3 equals negative 7 cosine of theta. At this point, I need to combine like terms. 00:41:00.885 --> 00:41:10.484 I need to take the negative 7 cosine of theta to the left side, set the whole equation equal to 0. 00:41:10.486 --> 00:41:28.417 2 cosine of theta squared plus 7 cosine of theta, a negative 1, a negative 3, gives me negative 4. I have the terms 00:41:28.419 --> 00:41:45.418 in descending order, so I can factor. This factoring, I'm not sure what to do. Let me let u equal the cosine 00:41:45.420 --> 00:42:05.168 of theta. 2u squared plus 7u minus 4 equals 0. I know the factors of 2u squared are 2u and u, and I think 00:42:05.170 --> 00:42:18.167 the factors that I'm going to need for 4 are 4 and 1 with a plus sign here, minus sign there. Double check. 00:42:18.169 --> 00:42:31.669 On the outside I have 8u. On the inside I have negative 1u. That combines to positive 7u. Let's just make 00:42:31.671 --> 00:42:49.666 the substitution now with cosine theta. 2 cosine theta minus 1 times cosine theta plus 4. Once I have the 00:42:49.668 --> 00:43:06.208 product of these two factors set equal to 0, I can set each factor equal to 0 and solve. 00:43:06.210 --> 00:43:25.418 2 cosine theta minus 1 equals 0. I'm going to add 1 and divide by 2. Cosine theta plus 4, I simply subtract the 4. 00:43:25.420 --> 00:43:36.417 Cosine of theta equals 1/2. This is a positive ratio, so once I find the reference angle, I know that will be one 00:43:36.419 --> 00:43:46.383 solution to this equation on my given interval. But let's find the reference angle, so we can find the second 00:43:46.385 --> 00:44:03.633 solution. Cosine of theta r equals positive 1/2. Setting up our triangle, I know that the cosine function is the 00:44:03.635 --> 00:44:15.157 ratio of the adjacent side, I place the 1, divided by the hypotenuse, so I place the 2, and I know that the opposite 00:44:15.159 --> 00:44:29.909 side is the square root of 3. So the reference angle is pi over 3. Ok, where else is cosine function positive? 00:44:29.911 --> 00:44:45.409 Quadrant four. That angle we will find by subtracting the reference angle from 2 pi which gives us 00:44:45.411 --> 00:45:02.408 5 pi over 3. So these are, the pi over 3, the 5 pi over 3, are the first two solutions. Now, let's go to cosine 00:45:02.410 --> 00:45:17.959 theta equals negative 4. I know it's not pi over 3, pi over 4, pi over 6, so let me go to my graph of my 00:45:17.961 --> 00:45:39.709 cosine function, y equals cosine x. Maybe I will find solutions there. Here we go. From 0 to 2 pi. 00:45:39.711 --> 00:45:50.209 Oh, I see the problem now. The graph doesn't reach negative 4. We know that the range of this function 00:45:50.211 --> 00:46:02.959 is from negative 1 to 1. So this equation gives us no solutions on our given interval, or ever really. 00:46:02.961 --> 00:46:14.210 So we have two solutions, pi over 3, 5 pi over 3, on the interval from 0 to 2 pi. 00:46:15.900 --> 00:46:25.209 Now we want to solve a trigonometric equation using a calculator. Our directions are to approximate all 00:46:25.211 --> 00:46:32.959 solutions to the equation on the interval from 0 to 2 pi, and we're going to round our answers to four 00:46:32.961 --> 00:46:43.650 decimal places. Here is our example. Sine of theta is 78 hundredths. Here's the calculator. The first thing 00:46:43.652 --> 00:46:52.493 we need to know about the calculator is that we put it in radian mode. Make sure you know how to do that. 00:46:52.495 --> 00:47:10.429 Ok, got that. So let's write that in our notes: radian mode. This ratio is positive, so we want to first find 00:47:10.431 --> 00:47:30.358 the reference angle, and we're going to do that by entering in our calculator sine inverse of this positive ratio. 00:47:30.360 --> 00:47:47.858 So we're going to do second, sine, then .78, enter, and we get that the reference angle is approximately, 00:47:47.860 --> 00:48:03.974 let's change that, approximately .8947. Ok, I've rounded this 7. It was 6 6, and then I've rounded to 7. 00:48:03.976 --> 00:48:14.975 The ratio is positive so that will give us one solution, our reference angle. The second solution will be found 00:48:14.977 --> 00:48:28.727 in quadrant two, which we will find, let's call it theta two, by subtracting the reference angle from pi. 00:48:28.729 --> 00:49:01.723 So let's write our solutions: 0.8947, pi minus that number which according to our calculator is 2.2469. 00:49:01.725 --> 00:49:11.726 Those are the two solutions for the trig equation sine of theta equals 78 hundredths.