 Welcome back to our lecture series math 1060 trigonometry for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misildine. In lecture 21, we're going to start a brand new chapter, chapter 7, which is going to be all about trigonometric equations. We're going to explore the solution sets of equations involving trigonometric functions like sine, cosine, tangent, et cetera. Specifically in this video, we're going to investigate trigonometric equations that in some regard can be viewed as linear equations. We'll look at some more complicated equations a little bit later. Consider the equation 2 times sine of x minus 1 is equal to 0. Can we find all values of x for which this equation would be solved, any choice of x that can we put in there? Well, one way to solve equations is graphically. You could take the function f of x equals 2x, 2 sine of x minus 1. You could take that and then we have to then find out all the places where f of x equals 0. That is, we'd be looking for the x intercepts of the function 2 sine of x minus 1. Now, be aware that's not this graph right here. That'll come into play in just a second. If we could graph the function, which we could graph something like this, we absolutely could. We could then find the solutions graphically. But it turns out we'll be much more efficient if we try to solve this thing algebraically. What I'm going to do is the following. If you were to just slightly modify this equation, let's say we had something like 2y minus 1 equals 0, and then you're asked to solve for y. If we did that, we probably would be in a really good situation. It's like, okay, well, the first thing I would do is I would add one to both sides. We end up with 2y equals 1, and then we divide both sides by 2. This is so peaceful. I'm imagining Daisy's in a meadow right now. So I get y equals 1 half. We solve for y. But maybe you've noticed the little trick I did here is just I just did the substitution y equals sine of x. Because if we think about this original equation, 2 sine of x minus 1 equals 0, the exact same steps apply here. Add one to both sides. We then get 2 sine of x is equal to 1. We're going to divide both sides by 2. And so then we end up with sine of x equals 1 half. So solving this equation is basically just like solving any linear equation, such as this linear equation 2y minus 1 equals 0. The only difference is that we end with this trigonometric expression sine of x equals 1 half. And so now this is where this graph comes into play here, because you'll notice this graph right here is y equals sine of x. None of the transformations involved right here. When does sine of x equal 1 half? So really, we're trying to figure out if we think of this as a function where y equals sine of x. Notice the little mnemonic we were doing there, right? y equals sine of x. We're looking for where on the graph of sine does the y coordinate equal 1 half. And this is going to happen, of course, at pi 6th in the first quadrant and at 5 pi 6th in the second quadrant. If you prefer to think of it in terms of the unit circle, that's perfectly fine as well. Think of your unit circle right here. So you're going to have the angle of 30 degrees or pi 6th right here. And then in the second quadrant, you'll get 5 pi 6th, which references to 30 degrees or references to pi 6th in the second quadrant. In the first quadrant, both x and y are positive. In the second quadrant, you can x is negative but y is positive. In the third quadrant, it's negative negative. And then in the fourth quadrant, it's positive negative. So where is the y coordinate? Where is sine going to be positive? That, of course, happens in the first quadrant, the second quadrant. And so we get these numbers, pi 6th and 5 pi 6th. So then we could say our answer is going to be like, okay, x equals pi 6th and 5 pi 6th. But it turns out that's not exactly the whole solution set either. This would be the solution set if x was between 2 pi and 0, right? If you were to think about the unit circle, if you went around, if you went around one whole time, you could end up with 2 pi plus pi 6th. That would also be a solution. You could take 2 pi plus 5 pi 6th. That would also be a solution. And so anything coterminal 2 pi 6th or 5 pi 6th would also be a acceptable solution. So when you often see when it comes to solving trigonometric equations, we're gonna give you the reference angles basically for the general solution. So we'll get things like pi 6th plus 2 pi k where k here is understood to be any integer, right? So you could add any integer multiple of 2 pi to pi 6th and that would be a solution. This includes where k equals 0 and therefore you add nothing, you just have the original pi 6th. And so x equals pi over 6 plus 2 pi k. And then we also would say 5 pi over 6 plus 2 pi k. This would likewise be a solution here. And so this right here gives us the general solution to this linear trigonometric equation. We have to find the, basically we have to find the solutions within one rotation of the unit circle. Basically we want to find all solutions from 0 to 2 pi. And then from there, we can add multiples of 2 pi from these to get all of the solutions. But in particular, within one rotation of the unit circle, there are in fact two solutions. One in the first quadrant right here, one in the second quadrant. And so we have to recognize well sine is positive one half, where is that obtained? Now in this example, because sine was equal to one half, that's a ratio we know very well coincides with 30 degrees or pi 6 right here. We're able to do this without the need of a calculator. Of course, if the sine ratio is not one of those special ratios, we might need to use inverse sine to help us compute this. And we'll see an example of that shortly. Consider now the equation two sine theta minus three equals zero. Let's solve for theta here. We're going to do what we did on the previous one. Solve for sine of theta first. So we're going to add three to both sides. This is then going to give us two sine theta is equal to three. Divide both sides by two. This algebraic part of the process is fairly simple just solving a linear equation. And so then we end up with sine theta equals three halves. Now you might be thinking, well, is there any place on the unit circle where sine theta equals three halves? No, there's not. Maybe you use sine inverse at this moment. If you did pull out your calculator and you asked it to compute sine inverse of three halves, you would get an error at this moment. And why is that? Remember that sine, this is also true for cosine. Sine is bounded by one and negative one. The range of sine and again, the range of cosine is one to negative one. You cannot get any value, any ratio for sine that's larger than one or cosine, nor can you get a ratio for sine or cosine that's less than negative one. Three halves, of course, is larger than one. So this actually gives us no solution in this situation. Some people like to draw this little circle of the line through it to be shorthand for no solution. That's perfectly acceptable in this situation. There's no solution. Let's think of this graphically. If you see the equation, the graph of the equation y equals sine of x right here. If we're looking for the y-coordinate of three halves, which would be up here, there's no place on the graph where sine is equal to three halves. So if you ever have to solve an equation for sine or cosine, which is outside of the interval negative one to one, there would be no solution in that situation. There's not multiple solutions in case there would be no solution. Let's consider another example. Let's find all of the solutions to the equation cosine of a minus 25 degrees is equal to negative root two over two, okay? Now with the previous equations, it didn't actually specify how should we express the value x? Should it be expressed in degrees? Should it be expressed in radians? It really ought to tell you that. In a trigonometry course and other precalculus and calculus courses, if the angle measure is not specified, the units that is by default should assume radians. This one does specifically tell us degrees, but also since we'd see an angle measurement described in degrees, that's a Q to us that, okay, the answer should be described in degrees. So how are we gonna do this one? Well, if you think about this just for a moment, let's just think of this as theta, right? We're trying to solve the equation cosine of theta is equal to negative root two over two, okay? And so in terms of the unit circle, right? Where would you get negative root two over two? Will the standard angle, just think of root two over two, right? That would happen at a 45 degree angle in quadrant one. That's normally where you get the point root two over two for X and for Y, right? But it's a negative root two over two. So where is cosine gonna be negative? That happens when you're to the left of the Y axis. So you're gonna get this one right here. So this is, you have a reference angle of 45 degrees and this one right here, where can you get a reference angle of 45 degrees? So this of course is gonna, this the one in the second quadrant, this will be 180 take away 45 degrees. That's of course going to give you 135 degrees. And then this other one in the third quadrant right here, this would be 180 plus 45 degrees. So that's 225 degrees. These are gonna be then the reference angles for the general solution. That is to say, we get that theta equals, we got 135 degrees plus two pi K, not two pi, we're in degrees. So we should say 360 degrees K. And then your other solution would be 225 degrees plus 360 degrees K, like so. But then we saw for theta, theta was just this intermediary variable there. Really we have A minus 25 degrees equals 135 plus 360, 360 K and then 225 plus 360 K. How do you solve for A in this situation? Well, it's the same as any other linear equation. You're gonna add 25 degrees. You'd add 25 degrees to this one right here. You'd add 25 degrees to this one right here. And so we see that the general solution would look like A equals, you're gonna take 135 plus 25, which is 160 degrees. You're still gonna get anything coterminal to 160 degree would also be a solution. So you'll still add in that 360 K. And then you're gonna take 225 plus 25, which is 250 degrees. Again, anything coterminal to that would also be a solution. So the general solutions are gonna be 160 and 250 or anything coterminal to that one. Let's look at one last example of solving a linear trigonometric equation. In this situation, notice how you have a sign here and a sign there. That's not gonna be too much of an obstruction for us because if we think of it as, oh, okay, we have three Y minus two equals seven Y minus one. What would you do? Well, you're just gonna try to combine like terms. We'd be like, okay, I'm gonna subtract three Y from both sides. I'm gonna add one to both sides. So we end up with a negative two plus one is a negative one. And then we get seven Y minus three Y is four Y. I would then divide both sides by four and we end up with Y equals negative one fourth. That's how you would get there. So what changes in this situation? Really nothing. Nothing's gonna change whatsoever. You have three sine theta minus two equals seven sine theta minus one. So we're gonna do the exact same thing. We're gonna subtract three sine theta from both sides. We're gonna add one to both sides. We end up then with a negative one is equal to four sine theta divide both sides by four and you end up with sine theta is equal to negative one fourth. Which notice there, negative one fourth does fall within the range of negative one and one. So there are gonna be solutions. And in fact, we should expect two solutions here. Well, where is sine gonna be negative? If we think of our unit circle diagram, something like this, right? If sine is negative, that means you're gonna expect a solution in the third quadrant and in the fourth quadrant right here. It does want solutions within zero to 360 interval that's specified to us. So we don't have to worry about any of the 360K or anything like that. Cause we just want just the reference angles to the general solution here. Now negative one fourth, if you think of it as positive one fourth either, it's like that's not one of those angles I was asked to memorize for the unit circle. All right, so what you're gonna do is basically you're gonna look for sine inverse of negative one fourth. That's what we do right here. Theta equals sine inverse of negative one fourth. You can consult a calculator with this one and with the calculator what it's gonna tell you, again, if you throw this into the calculator you're gonna end up in the negative 14.5 degrees making sure it calculates in degree mode cause that's how we wanna report the final answer. Well negative 14 degrees, negative 14.5 degrees, this would tell you that if you went counterclockwise from the positive X axis, 14.5 degrees that's gonna give you one angle for which sine is equal to negative one fourth. Now we need our solutions to be in the interval zero to 360. So that 14.5 degrees is the reference angle we wanna do. We're looking for the angle which is 14 degrees, 14.5 degrees below the positive X axis and we're looking for the angle which is 14.5 degrees below the negative X axis as well. So to think about that, this one over here you're gonna take 180 degrees plus 14.5 that's how we get this one. And then this one over here we're gonna take 360 minus 14.5. When you use your calculator to compute inverse trigonometric functions they only give you the reference angles. It's important to remember it can be the reference angle it doesn't give you all the solutions. Now of course, if you're in the first quadrant reference angles and the angle you get are the exact same thing but as we're in the third and fourth quadrant the reference angles give you something very different. So we have to be careful about that. Therefore for theta we end up getting 180 plus 14.5 degrees that's gonna be 194.5 degrees. And then the second one we take 360 take away 14.5 degrees that gives us 345.5 degrees as our solution. So these would be the two solutions between zero degrees and 360 degrees. If we wanted all the solutions to this equation we could take 194.5 plus 360 k and we take 345.5 plus 360 k like so. Again, if we want the general general solution but this equation only asked us to find the solution from zero degrees to 360 degrees. So when the angle or should say when the ratio coincides to one of the special angles you can get away without having to use these inverse trigonometric functions but in a case like this one you have negative one fourth no one has that memorized. So use the calculated compute sine inverse of negative one fourth but remember this right here is always gonna be the reference angle, right? And so that's not the answer but it does help you determine the answer and think of the unit circle to help you as you move from reference angles to the genuine solution to these trigonometric equations.