 Let's talk about how to solve trig equations. An equation is a trig equation. If the variable you're solving for is located within a trig function. For instance, in my first example, part A here, I have two sine theta plus square root of three equals zero. I wanna find the solutions between zero and two pi because there's infinitely many solutions, so to speak. Only do the ones between zero and two pi. So first thing I wanna do is I want to isolate the trig function. So subtract square root of three from both sides, then divide by two on both sides. I'm trying to find out sine of what angle is negative square root of three over two. First off, what quadrants is sine negative? That would be quadrants three and four. So all right, sine of sine is negative theta is in. Quadrant three and four. So the quadrant three and four. So now I'm gonna draw my triangle and I'm not gonna put my angle theta here in the bottom left. This is going to be my reference angle, my gamma, or as I like to put a question mark. Sine of this angle is square root of three over two. Opposite is square root of three, hypotenuse is two, meaning my other side has to be one. So what angle is always across from square root of three? What angle is always across from square root of three? That's 60 degrees, that's pi over three. That's my reference angle and I have to move this angle to quadrant three and move this angle to quadrant four. Those will be my solutions data to this equation. So I use my formulas I previously learned to move an angle to quadrant three. You take the reference angle and you add pi to it. Pi over three plus pi is four pi over three. To move an angle to quadrant four, and in this case, I want positive quadrant four. I take two pi and I subtract my reference angle pi over three. So that is six pi over three minus pi over three, which is pi pi over three. So I have two solutions to this trig equation, four pi over three and five pi over three. Those are my two solutions in the interval zero to pi. Next, this one looks a little bit different because instead of just having theta on the inside of the trig equation, we have three theta over two. So one way to get around this is, okay, well, let me pretend like I just have one given variable value inside of my trig function. So three theta over two, let's just call it data. So everything's going to be in terms of data right now. Then we'll convert back to three theta over two. In this example, I'm trying to figure out where secant is two over one. First of all, the secant is negative. My inside of my trig function beta, which is three theta over two is in quadrants two and three. Because that is where secant and cosine are both negative. Guess we can go ahead and draw our triangle. We have our mystery angle. This represents my reference angle. Secant is two over one. So if cosine's adjacent over hypotenuse, this is hypotenuse over adjacent. And lucky here, we're trying to figure out who is always across from square root of three. Well, that would be pi over three. Now I need to move this guy to quadrants two and three. So first off, let's move to quadrant two. So my quadrant two angle, which remember we're using beta here instead of all this theta mess that's inside the trig function. Beta is pi minus pi over three, which is two pi over three. And then moving to quadrant three, beta is pi over three plus pi. So beta is four pi over three. But remember this original question was not in terms of beta. It was in terms of theta, theta, three theta over two. I need to find out what theta equals. Well, beta is three theta over two. So substitute three theta over two back in for beta. I like to cross multiply three theta times three is nine theta, two times two pi is four pi. Dividing both sides by nine gives you theta is equal to four pi over nine. And lastly, over on the right-hand side, we have three theta over two because that's what beta is equal to. Three theta over two equals four pi over three. Cross multiply nine theta equals eight pi. Theta equals eight pi over nine. Now these are two of my solutions, but we have a trick here. And the trick is the period of my secant function is normally two pi. So if I had a normal secant function here with just theta on the inside, adding two pi would clearly get me out of the interval zero to two pi. But this trig function has a little bit different period. So the period is actually going to be two pi over whatever's inside the trig function, three halves in this case. So two pi over three halves, which is actually going to give you four pi over three. This is the period of my trig function here, four pi over three. It has been shrunk down from two pi. So what I should do since the trig function repeats itself every period, I need to add four pi over three to each of my angles theta, to each of my initial solutions and see if I'm still within the interval zero to two pi. So we add this to each value theta. So first off, I take four pi over nine and I'm going to add four pi over three to it. This would be four pi over nine plus 12 pi over nine, which gives you 16 pi over nine. This is less than two pi. It is, because 18 pi over nine would represent two pi. So this is less. So that we have found yet another solution, 16 pi over nine. If you were to add another four pi over three to this, you would definitely get over two pi. Let's consider the other solution, eight pi over nine. Let's add four pi over three to it. So I have eight pi over nine plus 12 pi over nine, gives me 20 pi over nine, which does exceed two pi. So this does not work. So it turns out we have a total of three solutions. I'll write them in order four pi over nine, eight pi over nine and 16 pi over nine. These are the solutions that are in the interval zero to two pi. So be careful on this one. You do need to make sure you add your period of your given trig function to each of your initial solutions because there could be a third solution lurking. So make sure you do that as if that one wasn't enough fun. Let's do yet another one. Step one, isolate the trig function. So add square root of three to both sides. And once again, rather than having just data inside the trig function, we have data over two. So if you want, replace data over two with a different Greek letter, we oftentimes like to just use something like beta. So theta over two is equal to beta. Now I'm gonna write square root of three over one because often like to have ratios. I guess we're ready to make us a triangle. So my original reference angle, which I represent with the question mark is here. So opposite over adjacent, square root of three, one, two, who is always across from square root of three? Well, if you haven't picked up on it yet, that would be your 60 degree angle here, pi over three. Now I gotta figure out what quadrant these solutions are allowed to be in. Well, since tangent is positive, since tangent is positive, where is tangent positive? Theta over two or beta would be in quadrants one or three. So this pi over three, this reference angle here, I need to move it to one. Well, guess what? It's already in one. So that's less work we have to do. But I do also need to move to three. I need to move to quadrant three. So my solutions include beta is pi over three, or replace it with theta over two in just a minute. Or beta is take your reference angle and add pi to it. This is how we move an angle to quadrant three. We add pi to it. So beta is four pi over three. Now remember, our answer needs to be given in terms of theta. So we write theta over two equals pi over three. We have three theta equals two pi. We have theta equals two pi over three. Now we also have theta over two equals four pi over three. We have three theta equals eight pi. We have theta equals eight pi over three. So our solutions so far are theta equals two pi over three, eight pi over three. But please be aware that because we had theta over two inside our trig function, we could have additional solutions between zero and two pi if we add the period of our trig function to our initial solutions for theta. Only way we're going to do that though is take our period and find it. Well, the period of tangent is pi divided by whatever is inside the trig function, which is one half in this case. So that would mean our period would be two pi, which gets us out of the interval, zero to two pi if added to any of the given trig solutions. So we only have two solutions here, two pi over three, eight pi over three. So far we've only been looking at trig solutions in the interval zero to two pi. But what if we want to find all the solutions, infinitely many solutions for our trig functions? So in this example, we're going to give the general form of the solution and then we're going to list six solutions. So sine of theta equals square root of two over two. I don't have any side relationship of a triangle that has square root of two and two in it. So this is something we need to reverse rationalize here and we get one over square root of two. Sine of theta equals one over square root of two. And a quick little note here, we write since sine is positive, the inside of our trig function theta is in quadrants one and two. So we're in quadrants one and two. So let's build us a triangle with our mystery reference angle inside here in the corner. Sine is one over square root of two, opposite over hypotenuse, which I mean the other side would have to be one. This is a 45, 45, 90 triangle. Yes, my reference angle is 45 degrees or pi over four. We're wanting to move it to one and four, one and two. It's already in one. So I just need to move to two. So theta is pi over four because that is the quadrant one angle and theta is, how do you move an angle to quadrant two, where you take pi and you subtract the reference angle, the angle that we found. So theta is three pi over four. So I received pi over four and three pi over four. Remember previously we had to take the period of the trig function and add it to these initial values found to find given solutions. And that's how we find the general form of the solutions for this trig equation. The period of sine theta is two pi. So to find general solutions, we take each solution of theta and add multiples of two pi. So we add multiples of two pi. So my first general solution for theta, I'll represent it as theta sub one would be pi over four plus two k pi. We know where the two pi came from and k is just a reference and integer. So for instance, when it's k is zero, you get the solution pi over four. When k is one, you add two pi to pi over four. When k is two, you add four pi to pi over four, all multiples of two pi. So this is plus two k pi where k is an integer. Then our second solution for theta is three pi over four plus two k pi. k is an integer. Now they want us to list six exact solutions. So, okay, first we got our general solutions here. Done, check. To find our six solutions, we'll plug in k equals zero, one, two. Cause this will give me, zero will give me two solutions. One gives me two solutions, two gives me two solutions. All right, so when k equals zero, we'll start off easy here. Theta one is just pi over four. Theta two is three pi over four. And k equals one, comma. Theta one is pi over four plus two pi, which is nine pi over four. And then theta two would be three pi over four plus two pi, which is 11 pi over four. Then k equals two. Theta one is pi over four plus four pi. Four pi, that's going to give you 17 pi over four. Theta two is three pi over four plus four pi, which is going to give you 19 pi over four. You may notice a pattern here, which results to patterns and trig. And k is zero, you had one pi over four, three pi over four. And k is one, you had nine pi over four, 11 pi over four. And k was two, you have 17 pi over four plus 19 pi over four. They all, all the numerators differed by just two for each of the values of k, which is a pattern that does happen here. So we had our solutions here. Six solutions will be three pi over four, pi over four, nine pi over four, 11 pi over four, 17 pi over four, and 19 pi over four. And these are six solutions that we obtained from the general solutions. And of course, there's infinitely many more, but these were the easiest six for us to find. So hopefully you learned a little bit about solving trig equations and I appreciate you watching.