 In this video, we provide the solution to question number 13 for practice exam number three from math 1060. And we're asked to find all of the solutions to the equation cosine of two theta plus five cosine theta plus four equals zero. And we want to do that on the interval where theta ranges from zero to two pi. Notice by the interval, we're looking for solutions that are in radians. And we need to solve this equation. Well, we do have a mismatch of angles here, we have a cosine of theta and a cosine of two theta. So we probably need to use the double angle identity on cosine in order to get just theta's here. But there are three versions of the double angle identity for cosine. There's cosine squared theta minus sine squared theta. There's two cosine squared theta minus one and there's one minus two sine squared. Did I say that one right? One minus two sine squared theta. There you go. Since we have a cosine theta right here, our strategy is to use the identity that cosine of two theta is equal to two cosine squared theta minus one. Because if we plug this into the equation, we then get a quadratic equation for cosine, which hopefully we can then solve that by factoring or the quadratic formula, whichever we prefer. So we're going to get two cosine squared theta minus one plus five cosine theta plus four equals zero. The thing is, if this was a five sine, we would have actually instead used the identity one minus two sine squared theta instead. We're using this version of cosine two theta because it's compatible with the cosine right there. Combined like terms, if possible, there's a negative one and a plus four, so we can add those together. We get a two cosine squared theta plus five cosine theta plus three, like so. And so if you put these together right, can I factor this? Two times three is equal to six. Six. Is there any type of factorization that I could utilize in that situation right there? Well, of course, if you just take two plus three, two times three again, two plus three is equal to five, is I guess what I'm trying to say there. We already have the magic pair in front of us. And so we can factor this by reverse foil. I bet I could probably guess the factorization because two and three are prime numbers, so there's not a lot that could go on there. To get a two cosine squared, you're going to have to have a two cosine theta and a cosine theta. That's the only option there. I want the two to go with a one, and I want the one to go with a three, right? Talking about the coefficients two and one makes a two, and three and one gives me a three, and then three plus two gives a five. So this is in fact the correct factorization. We could have gotten through more details with the group factorization, the reverse foil if we needed to, but we can guess it and that's okay. You can also use the quadratic formula if you prefer. So setting each of these factors equal to zero, you get two cosine theta plus three equals zero. Solving this one, you get a minus three, two cosine theta is equal to negative three. Cosine theta is equal to negative three over two. Now remember that cosine, just like sine, is bounded above by one. It's bounded below by negative one. Negative three halves, of course, is negative 1.5. So it turns out there's no solution in that case. Cosine cannot equal negative three halves. You could have also taken arc sine of negative 1.5. You would have gotten an error with your calculator, which indicates there was no solution there. So looking at the second factor here, there's a possibility there, cosine theta plus one equals zero. That would suggest that cosine theta equals negative one. And when does cosine equal negative one that happens on the very left of the unit circle that happens at theta equals pi. And we only need answers between zero and two pi. And so that turns out to be the only solution to this quadratic equation. The answer is going to be theta equals pi.