 In our lecture series, we've seen many instances where we had to solve a quadratic trigonometric equation where we had a quadratic polynomial in terms of cosine or a quadratic polynomial in terms of sine, all right? What if you have a mismatch though? Like here, you have four cosine squared of x plus four sine x minus five equals zero. It kind of feels like a quadratic term because you have your linear term, your constant term and your quadratic term right there, but there's this mismatch. You have cosine squared and a sine right there. If you had cosine squared and a cosine, that'd be great. If you had a sine squared and a cosine that'd be great too. So it's the mismatch. We have a cosine and a sine. What do you do with that? Well, we wanna switch from one to the other. And there are some options and basically the best options come from the Pythagorean relationship. Now, we don't wanna switch the linear term because the linear term would be something like the following sine of x is equal to plus or minus depending on the quadrant, the square root of one minus cosine squared which that does not reduce. Otherwise we would do it. Introducing square roots into the problem will be a big no-no yikes. Another possibility we could try would be to use like the half angle identity or something but no, no, no, that would even be worse. But the Pythagorean identity is where the money's at. We don't wanna switch the linear term. We wanna switch the quadratic term because the quadratic polynomial cosine squared plus sine squared equals one. We can solve for the quadratic term we have. So solve for cosine squared. We'll get cosine squared is equal to one minus sine squared like so and we can make this substitution in for cosine squared. Thus we would switch from cosine squared into a sine squared and that's the setting that we want. So replace the cosine squared with one minus sine squared like so. We keep the four sine. We keep the negative five of course equals zero. Notice that we're trying to solve this on the domain zero to two pi. So that clearly is telling us we wanna solve this equation for radians. Just one rotation of the unit circle here. Distribute the four. We're gonna get four minus four sine squared plus four sine of x minus five like so. Combine like terms. You end up with four sine squared x plus four sine x. And so the constants will combine here. You get four minus a five there. So you get a negative one equals zero. I'm going to divide both sides by negative one. I don't like the leading coefficient being negative. It just frustrates me a little bit. So maybe I'm just too much of an optimist there. But if you divide everything by negative one, you'll get four sine squared minus four sine x plus one now equals zero. Now we have to decide do we wanna factor this thing? Do we wanna use the quadratic formula? That decision is possible of course here. Now I noticed that the first term has a coefficient of four. The last one's a positive one. Those are both perfect squares, right? This is two squared and this is one squared. Which makes me think this possibly is a perfect square trinomial. Notice the middle term. Basically I'm utilizing the factorization here that a plus b squared equals a squared plus two a b plus b squared like so. Or in this case, I guess we're taking a minus b squared is equal to a squared minus two a b plus b squared like so. So does the middle term match up? So this should look like negative two times two times one times sine of x. That's exactly what we have right there. Two times two is four. So this does factor as a perfect square trinomial. So we're gonna end up with the factorization two sine x minus one quantity squared equals zero. If you didn't see that, if you tried the usual guess and check or reverse foil method, you'd end up with like two sine x minus one times two sine x minus one. The point is you'd get a repeat there. And so really there's only one case to consider. What happens when two sine x minus one equals zero? So we'll consider that exact situation. Add one to both sides. We get two sine x equals one. Divide both sides by two. We end up with sine of x equals one half. So sine is positive in the first quadrant and in the second quadrant. In the first quadrant, x would be pi six, they're 30 degrees, a bit we're solving with radians here. And then in the second quadrant, we need the angle that references pi six, which we pi take away pi six or five pi six, which would be of course the solution in this situation. So if you have a quadratic equation with trigonometric functions where there's a mismatch between sine and cosine, use the Pythagorean equation to convert from, to convert the quadratic term to be compatible with the linear term. And then you can solve this like any other quadratic trigonometric equation.