Solve this equation without a calculator

What's up, guys? One thing that trips up students a lot of times is when we need to evaluate an expression, and we have a restriction. Either we evaluate the restriction from 0 to 2 pi, or find all of the solutions that make this equation true. So in that video, that's exactly what I want to do. I want to find the values that make this equation true. And we're going to do that for a restriction, as well as for all the values. Now, one of the things I always like to tell my students is when we're trying to evaluate trigonometric expressions, one thing we always want to look for is what is the value that's on the unit circle? Because if we can evaluate or find the value that makes it on the unit circle, it makes our life a lot easier. And if you look at this, you might recognize that's not on the unit circle. I don't recognize this point on the unit circle. If you were to imagine the unit circle, visualize it, I don't remember that. However, there's something we can do here. We can rationalize the denominator. When rationalizing the denominator, you're just simplifying the expression. So now, and again, just by multiplying the square root of 2 on the top and the bottom, all I did is I just simplified this expression. So now, there's no longer a square root on the bottom. But what I also did is I created a value that I now recognize on the unit circle. So knowing your unit circle or knowing where these values is is paramount for you to be able to understand how to evaluate this equation. So there's a couple of things that I did really quickly in my head that I remembered. So the first thing is I always remember the first quadrant in the unit circle. You need to know that this angle is pi over 4. And on that point of the unit circle, for the angle pi over 4, or 45 degrees, you're going to have the coordinate point square root of 2 over 2, comma square root of 2 over 2. Now, again, that is your x, and that is your y-coordinate. So remember, this is the cosine of what angle is equal to square root of 2 over 2. Now, on the unit circle, which is basically the restriction of 0 to 2 pi, we could say that one of the angles is going to be a pi over 4. Now, what about the other angle? And why did I so quickly go over here? Well, the main thing is kind of knowing your signs of, on the unit circle. In the first quadrant, x and y are always going to be positive, right? If you go to the second quadrant, though, notice on this x-axis, we're going to the left of the y-axis. So therefore, all my x values are now going to be negative. So for the second and the third quadrant, if I was looking for a negative square root of 2 over 2, that's where I would have looked. So I know I have to go over here to square root of 2 over 2, comma negative square root of 2. Now, one mistake that students want to do is sometimes they'll do the negative version. They'll do negative pi over 4, right? Because you go pi over 4, and they're like, oh, negative pi over 4. Well, that all is going to depend on your restrictions. So on this first one, I'm saying I only want you to find the values between 0 and 2 pi, OK? That's going to be the restriction. So negative pi over 4 is a negative angle. That is not on this restriction of 0 to 2 pi. So what that means is we need to continue in the positive direction. So we can go and add these lines here. And if you're not already familiar with exactly what it is, then just do the simple thing and count. Pi over 4, 2 pi over 4, 3 pi over 4, 4 pi over 4, 5 pi over 4, 6 pi over 4, 7 pi over 4. So therefore, my other angle that satisfies this equation is going to be a 7 pi over 4. Now, those are going to be the two angles within the restriction of 0 to 2 pi. And that's usually one of the most common answers that we will expect students to be able to know. But there's another thing that we need to be able to figure out. And that is going to be all of the solutions, OK? So when we're trying to find all the solutions, one thing we want to do is look to see if there's any relationship that we can find between our two solutions, right? So here's one solution that we found. And here's our other solution that we found. Now, you can see that they are pi halves away from each other, right? Pi over 4 plus pi over 4 is 2 pi over 4, which is pi halves. But if I add another pi halves, I don't get to a solution, right? So there's no pattern that we basically have. So the best way for us to be able to find additional solutions to take pi over 4 is if we add another revolution, right? If we take pi over 4 and go over 2 pi, well, that's all the way around the circle. So you go add 2 pi, you're back over here. Add 2 pi again, add 2 pi. And also, you could subtract 2 pi, subtract 2 pi, subtract 2 pi, subtract 2 pi. So therefore, we want to find all of the solutions. We could say x is going to equal a pi over 4 plus a 2 pi. Now, how many times are we going to have to add 2 pi? Well, we don't know. It's infinite many times, right? We can go in the positive direction or the negative direction. So we're saying 2 pi n, where n is going to represent any integer. And we're going to have to do the exact same thing here for a 7 pi over 4 because, again, there's no relationship between those two. And I'll get to one of those examples here coming up. So 7 pi over 4 plus 2 pi n. So that is going to be all the solutions. These are the solutions between 0 and 2 pi. And that is how you evaluate that expression. Hopefully, this video was helpful for you and gave you some value. And if it did, I know you're going to love the next video I have for you here.