 In the previous video, we talked about solving linear trigonometric equations. In this video, I want to talk about solving quadratic linear equations. When it comes to solving quadratic equations, so you have something of the form ax squared plus bx plus c equals 0, you want to solve this equation or something equivalent to this, there's basically three strategies you learn in an algebra class that you can utilize in this situation. The first technique would be to factor it. There are various factoring techniques you could use to factor quadratic polynomials. Those factoring polynomials would also apply in a setting like this. 2 cosine squared t minus 9 cosine t equals 5. We'll see how to do that in just a second. The second strategy is to complete the square. Completing the square is introduced as a strategy because factoring is not always successful. Not every quadratic can be factored using just real coefficients or a whole number or a fraction coefficients as well. Frankly speaking, an algebra student never likes to complete the square. It's a really good technique it is, but let's be honest, no one ever likes to do it. We're just going to move on to the third option, which is the quadratic formula. The quadratic formula also solves any quadratic equation. You don't have to do any guess or checking because, like you do with factoring, the quadratic formula is mechanical, you can run through it. The arithmetic can get a little bit more complicated than factoring, but the quadratic format is an option. I should mention that the quadratic formula is derived by completing the square in the most general setting possible, aka this setting where we don't know the coefficients. Every time you use the quadratic formula, you really are completing the square, but you don't have to remember the algorithm for completing the square, you can just use the formula. I'm going to show you two examples here. This first one I want to show you how to solve this quadratic trigonometric equation using factoring, and then on the next one we'll use the quadratic formula, all in the same video here. So let's solve the equation 2 cosine squared t minus 9 cosine t equals 5. Now this can sometimes feel overwhelming, and if necessary, make the substitution y equals cosine of t. Because if you do that, then this equation looks like 2y squared minus 9y is equal to 5. And so without having to worry about the trigonometric expression for a little bit, let's just deal with this equation. When it comes to factoring, you want to set the right hand side equal to 0. So you end up with 2y squared minus 9y minus 5 equals 0. So then we have to combine together the leading coefficient, that is the coefficient of the squared term, and then the constant coefficient, multiply those and you get 2 times negative 5, which is equal to negative 10. All right? This, then we need to find factors of negative 10 that add up to be negative 9. And this one that's a pretty easy guess there, you're going to take negative 10 and 1. Notice that negative 10 times 1 of course is negative 10, but negative 10 plus 1 is negative 9. So this is what we have found here, a so-called magic pair, a pair of numbers which allows to reverse the foil process going on here. So then what we're going to do is we're going to replace the negative 9 actually with a negative 10y and a 1 there. So it's like the negative 9 became negative 10 plus 1 y minus 5 equals 0. Distribute the negative, excuse me, distribute the y here. So we get 2y squared minus 10y, and then you're going to get a y minus 5 is equals 0. So the whole idea of finding a magic pair so you can break the middle coefficient up into two terms so then you can proceed to factor by groups. So we'll put the first 2 in a group, 2y squared minus 10y, put the second 2 in a group. It doesn't matter, it doesn't matter how you do this. If you did the other order 1 then negative 10, it's going to work out just to be fine. And so in each of the individual groups you want to factor out the greatest common divisor. So between 2y squared minus 10y, you can factor out a 2y that leaves behind y minus 5. In the second group, the greatest common divisor is just a 1. So you have to factor out something even if it's just a 1. And so you're left behind y minus 5. You'll notice now they have a y minus 5, a y minus 5, those are the same term. You can factor out the common divisor there even if it's a binomial. And you end up with now 2y plus 1 times y minus 5. This is equal to 0. Now I'm going to then return to our original setting right because it kind of got off the screen right here. But you'll notice that we made the original substitution that y equals cosine of t. So now let's put y back into the equation. We're looking at the equation 2 cosine of t plus 1 times cosine of t minus 5 equals 0. So we factored the quadratic trigonometric polynomial in terms of these. And we just used y just as a helper variable so we didn't get confused with the cosine of t. We just think of it symbolically. Okay, y is cosine of t. Let's factor in terms of y. Now that we have this thing factored, we can use the zero product property which tells us that the only way a product of real numbers is equal to 0 is if one of the factors is equal to 0. So this suggests two possibilities. Either 2 cosine t plus 1 is equal to 0 or cosine t minus 5 is equal to 0. Let me rewrite that 5. 5 equals 0. Now I'm not saying t would simultaneously do these things but one of these possibilities has to happen. Now if you pursue the first possibility, we'll subtract 1 from both sides. You get 2 cosine of t equals negative 1. Divide both sides by 2. You get cosine of t is equal to negative 1 half. And so when is cosine equal to 1 negative 1 half? Well, if it's equal to positive 1 half in the first quadrant, do recall that the original instructions told us to find all solutions in the interval 0 to 2 pi. So we are going to be doing this in radians. If we were looking for cosine equals 1 half, again in the first quadrant, that happens of course when the angle is equal to pi thirds, 60 degrees. But we're having a negative right here. So this means our quadrant would actually be the second quadrant or the third quadrant. So what angles reference pi thirds in the second and third quadrant? Well in the second quadrant, this is going to be pi minus pi thirds, so 2 pi thirds. And in the third quadrant, this would be pi plus pi thirds. So you end up with 4 pi thirds right there. So this gives us two solutions in the domain 0 to 2 pi. What about the other one? Well if we try to solve this one, cosine of t equals 5. Notice that 5 is outside the range of cosine. Cosine has to be between 1 and negative 1. So there actually gets you no solution in this case. So this just means, that doesn't mean that the equation has no solution. It just means that this option turns out to not be valid. And so there does turn out to be only two solutions in the domain 0 to 2 pi. You're going to get the solutions t equals 2 pi thirds and 4 pi thirds. Now of course, if we wanted the general solution set, we would add 2 pi k to these things right here. But again, the instructions only ask us to find the solution from 0 to 2 pi. So we can solve a quadratic trigonometric equation by factoring, if we can of course factor the underlying quadratic equation. Let's try this again. This time we have 2 sine squared theta plus 2 sine theta minus 1 equals 0. And again, we want to solve this on the domain 0 to 2 pi. Well if we do again a helpful substitution, y equals sine of theta, this polynomial would look like 2 y squared plus 2 y minus 1 equals 0. We could try to factor this one. But when you try to factor, you're actually going to get some problems here, right? So if you take 2 and negative 1 together, you get 2 times negative, excuse me, 2 times negative 1, which is negative 2. Are there factors of negative 2 that add up to be 2? And the answer is basically no. I mean the only options you have are 2 and negative 1, which gives you 1. And you have negative 2 and 1, which gives you negative 1. That's the only way to factor negative 2. So it turns out factoring is not going to work here. So instead, we're going to use the quadratic formula. And so do recall how the quadratic formula works here. If you're solving for y, you get y equals negative b plus or minus the square root of b squared minus 4ac all over 2a. And if you have a hard time memorizing the quadratic formula, you could say that exact same formula to the tune pop goes the weasel y equals negative b plus or minus the square root of b squared minus 4ac all over 2a. Where, of course, what's a, b, and c? We're talking about the most general polynomial here, ay squared plus by plus c equals 0. So a is the coefficient of the quadratic term, b is the coefficient of the linear term, and then c is the constant coefficient here. So if we apply that to this polynomial, we would see that y equals, well, my b here is a 2. So you're going to get negative 2 plus or minus the square root of b squared, which is 4. Then you're going to get minus 4 times a here is a 2 times c, which is a negative 1 all over 2 times a, which a was 2. Like so we want to simplify this thing. Now some other things to note here, y is equal to sine theta. So we're getting that y, which is sine theta is equal to all of this. So we get negative 2 plus or minus. Let's simplify the discriminant here, the number inside of the square root of the quadratic formula. So you get 4. Notice here, we're going to get negative 4 times negative 1. That's equal to positive 4 times 2. So that's an 8. The denominator here looks like a 4. So we get negative 2 plus or minus the square root of 12 over 4. It might be tempting to try to cancel out a common factor of 2 there, but no. We can only cancel divisor of 2 in the denominator if the entire numerator is divisible by 2, which it is, but we don't see that yet. To see that, in fact, we're going to simplify the square root. 12 itself is not a perfect square. The fact that it's not a perfect square, the discriminant is why we couldn't factor this thing earlier. A magic pair only exists if the discriminant is a perfect square here. But 12 does have a perfect square divisor. 12 is 4 times 3. And if you take the square root of 4, you end up with 2. So we then get sine theta is equal to negative 2 plus or minus the square root, sorry, plus or minus 2 times the square root of 3 all over 4. And so now you can factor a 2 out from the numerator. So you get negative 1 plus or minus square root of 3 over 4. And so then now this factor of 2 on the top cancels with a factor of 2 in the denominator. And so we get our expression negative 1 plus or minus the square root of 3 over 2. So let's consider these answers for a moment, right? How big are these numbers, right? Because the thing to always remember here is that sine of theta must be bounded by 1 and negative 1. If we have answers outside of that range, that's going to be a problem for us. So notice the following. Like you could throw this in a calculator. You could take negative 1 plus square root of 3 over 2, or you could take negative 1 minus square root of 3 over 2. That's perfectly fine. But I want you to think about the following. If you take negative 1 plus the square root of 3 over 2, notice here that the square root of 3 is less than the square root of 4 because the square root is an increasing function. So if you replace 3 with 4, that would make the thing get bigger. The advantage of this, of course, is that 4 is a perfect square. So you get negative 1 plus 2 over 2. So you're going to end up with 1 half, which notice is less than 1. So notice here that negative 1 plus the square root of 3 is, of course, it's less than 1 right here. It's also a positive value. Square root of 3 is bigger than 1 there. So we see that, okay, negative 1 plus square root of 3 over 2. That is inside the domain. That one checks out. What if we go in the other direction? What if we were to take, well, I guess, again, let's look at the smaller value, right? This thing would be larger than negative 1 plus the square root of 1 over 2, which is equal to negative 1 plus 1 over 2, which is equal to 0, which, of course, is greater than equal to negative 1. So that was just finishing what I said a moment ago. Let's just be explicit and write it on the screen. So negative 1 plus square root of 3 over 2, that one falls within the acceptable range of sign. So that's going to lead to a legitimate solution. So if we're trying to consider this case here, sine theta equals negative 1 plus the square root of 3 over 2, well, then we would consult our calculator and we're going to say that theta equals sine inverse of negative 1 plus the square root of 3 over 2, right? This is going to be a positive value and we're looking for solutions in the first quadrant and in the second quadrant. If we consult our calculator here, well, again, if we want the exact answers, right, there's going to be this value right here and then the other solution would be pi minus sine inverse of negative 1 plus square root of 3 over 2. If you want the exact answers, that's what they are. We will approximate them, of course. The first answer that is the square root of negative 1 plus, excuse me, sine inverse of negative 1 plus square root of 3 over 2, that's going to be approximately 0.37 radians. The second one, pi minus sine inverse of negative 1 plus square root of 3 over 2, that'll give you approximately 2.77 if we round to 2 decimal places there. That's one possibility. What about the other possibility? Let's consider this one, negative 1 minus square root of 3 over 2, okay? If we play that game, right? If you replace negative square root of 3 with 4, right? How does that compare? Well, since this is a negative number, if you increase the radicand, that actually means you're subtracting more, which means this will be smaller than the other side. That is, you get something like this, okay? For which we try to simplify this, you get negative 1 minus 2 over 2, right? Which gives you negative 3 over 2, which that is outside the range, but we're saying that our number is bigger than that. So that's not necessarily a problem. So it turns out that's not crazy useful for us. But if we go the other direction, okay? If you take negative 1 minus the square root of 1 over 2, this gives us negative 1 minus 1 over 2. That is, this gives us negative 1. Notice how our number is less than negative 1, right? This tells us that this actually gives no option here. Sine cannot equal negative 1 minus square root of 3 over 2. You could also throw this in a calculator and see that negative 1 minus square root of 3 over 2 is less than negative 1. Therefore, there's no possible solution there. So in the end, when you're solving this quadratic equation using the quadratic formula, we get there's two solutions in the interval, right? In the interval that theta is between 2 pi and 0, okay? You get 0.37 and 2.77. So the quadratic equation gave you two solutions, but when you consider those two cases, one of them was successful and gave you two radiant solutions. The other one was unsuccessful, and that's not always the case. You could have, when you have you solve these trigometric quadratic equations, it could be that both cases are successful, so you have lots of solutions. It could be that like in this one, one works, the other one doesn't, or it could be that none of them work. That is also a possibility. So to solve these trigometric equations algebraically, we're combining the algebraic techniques of solving quadratic equations like factoring the quadratic formula, and then we're combining with our knowledge of trigometric functions that help us find these multiple solutions like 0.37 radians and 2.77 radians.