- Hi, everyone. Welcome back to another precalculus video. Today we're going to be looking at how to solve trig functions by factoring. 

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Now, as you're going through that, if you haven't watched my first two videos on isolating trig functions or using square roots, please watch those two videos first. But if you're ready to keep going in that, here we have factoring, so a little bit more of an advanced technique that's going to really help you solve a lot of these trig functions. 

So as we're going through this, you're going to want to bring all your trig functions to one side. So let's see what we can do here. I can do 2 sine of x, cosine of x. And I can subtract a square root of 2, cosine of x on both sides, so minus square root of 2, cosine of x. And we're setting it equal to 0. 

Now, it may not stay like that, but that's fine. All right? And now what I can do here is what I notice between both of these. So let me highlight that real quick. So I have two terms. My first term and my second term I've underlined separately. 

So between both of those terms they have one thing in common. And they both share a cosine of x. So I'm going to factor that out from both of those-- so cosine of x. So I factored that out. And I have what's left over inside, 2 sine of x minus square root of 2. And that's equal to 0 now. 

So now that it's factored, there's nothing else that they had in common for me to factor out. So now that it's factored, I can solve both of those factors separately. What you're going to do is set them both equal to 0. 

So my first factor, cosine of x, I'm going to take that over here and go cosine of x equals 0. And then my second factor, 2 sine of x minus square root of 2, and set that one equal to 0 because they're both equal to 0. And we're going to solve for which values of x are going to get us 0 in either of these cases. And all of those are going to be possible solutions for our trig function on the interval from 0 to 2 pi. 

So I'm going to do it just like I did in the last two videos, isolating my trig function and solving. So in this case, I get x equals the inverse cosine of 0. And I want to know, when is cosine-- my x value is on the unit circle-- when are those going to be 0? Well, it's happening at pi/2 and 3 pi over 2. So those are two possible solutions. 

But I have a lot more over here in this other factor. So I'm going to solve for sine of x. In this case, I'm going to add square root of 2 to both sides. And I'm going to get 2 sine of x equals square root of 2. And I'm going to divide both sides by 2. So sine of x equals the square root of 2/2. 

Now I'm going to take the inverse sine. So x equals the inverse sine of the square root of 2/2. 2. And what values is going to be true for that? So that's when those are my y values, and they're both positive, so quadrant I and quadrant II. And so my values for this are going to be pi over 4 or 3 pi over 4. 

So I have four possible solutions in this specific case. I got pi over 2 and 3 pi over 2, and I also have pi/4 and 3 pi over 4. And again, we're looking only on the unit circle from 0 to 2 pi. But one thing you do want to continue to keep in mind is that these functions have a different domain, especially for those inverse cosines and inverse sines. Those have a special domain that you're going to want to follow in the future. 

Now over here on the second problem, we actually kind of have a special look at this one, one that we haven't really seen before but one that you can solve by factoring just in a different way. In our last problem, we used the greatest common factor. What did they have in common? In this one, we actually have a quadratic trinomial. 

So what I'm going to do first is I'm going to let u equal cosine squared of x. And I'm going to make that substitution to make that a little bit easier on us. So if u is cosine squared of x, I actually get u squared in place of this cosine to the fourth of x. And since u is cosine squared of x, so in this case, I'm just going to get plus u, and then nothing to substitute for the minus 2 equals 0. 

And what I want to do is I'm going to want to factor that quadratic trinomial. Now, we've had a lot of practice factoring quadratic trinomials. So what we're going to do is we're going to factor this. And we're going to get u plus 2 and u minus 1 equals 0. 

And we're going to solve for both of those factors. So I get u plus 2 equals 0 and u minus 1 equals 0. I'm going to solve for u in both cases. So u equals negative 2, and u equals positive 1. 

Now, we made a substitution for u. We said u was cosine squared of x, just to make it a little bit easier for us to work with. So now that we've gotten it down to here, just to solve for u, we want to bring that substitution back. So cosine squared of x is equal to negative 2. And cosine squared of x is equal to 1. 

So now you want to make sure and utilize what we did on our last video with using square roots. I'm going to take the square root of both sides, so square root and square root. And I'm going to get cosine of x equals-- and you're going to run into a problem in this case right here. 

You end up trying to take the square root of negative 2. Well, the square root of negative 2, you cannot take the square root of negative values and get real numbers. You end up with the imaginary numbers. And there are no imaginary numbers on the unit circle. 

So in this case, for this part of the problem, there's no part that we can keep going with this. There's no solution in this case. So we can ignore that part. But we still have this whole separate factor that we need to solve for. So let's get into that. 

We're going to take the square root of both sides. Square root and square root, and I get cosine of x equals-- and remember, when you take the square root, you get positive 4 and negative 1. And then we're going to get x equals the inverse cosine of positive or negative 1. 

So on the unit circle where my cosines are positive or negative 1, so where my cosines are positive or negative 1, that's going to be at 0 or pi, is going to be at positive 1 at 0 degrees or 0 radians, and it's going to be at negative 1 at pi radians-- so 0 or pi. 

So there are some different factoring techniques that are going to come into play here. It's going to take a lot of practice I'm not going to lie to y'all. So you're going to need to really practice these if you want to get good at them. All right? The pattern is going to be the same. You just need to recognize when to use which technique. 

So I have four more problems here for you guys to do. What I'm going to do here is I'm going to speed up the video. If you want, you can pause the video and give all four a try. Or you could just watch it till the end and see if you got the correct answers. 

All right, so I'm going to speed through the video so you can see all the work as it happens. So slow it down if you need. 

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- And 4 pi over 3-- all right, so that's going to conclude all four of those practice problems for y'all. There are a ton more practice problems on my website. So make sure you get to practicing. This last problem was probably pretty difficult for if you haven't seen factoring by grouping in a while. But if you have, it does work out very nicely. 

So no matter what factoring techniques, there are a ton of them. They're all going to be utilized here. If you're working through and you're struggling, please reach out to me for help. I'm Mr. Hernandez, and I'm always here to help. 

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