Cosine equation solution set in an interval

In a previous video, we established the entire solution set for the following equation. We saw that all the x's that can satisfy this equation are a combination of these x's and these x's. Here, the reason why I'm referring to each of them is numerous x's. For any integer value of n, you'll get another solution.

What I want to do in this video is to make things a little bit more concrete. The way that we're going to do it is by exploring all of the x values that satisfy this equation that sit in the closed interval from negative pi over 2 to 0. So I encourage you, like always, pause this video and have a go at it by yourself before we work through it together.

All right, now let's work through this together. The first helpful thing is we have these algebraic expressions. We have things written in terms of pi; let's approximate them all in terms of decimals. So even pi over 2, we can approximate that. Let's see if pi is approximately 3.14; half of that is approximately 1.57. So we could say this is approximately the closed interval from negative 1.57 to zero.

Negative 1.57 isn't exactly negative pi over 2, but it'll hopefully be suitable for what we're trying to do here. Now let's see if we can write the different parts of these expressions or at least approximate them as decimals. This could be rewritten as x is approximately, if you were to take 1/8 times the inverse cosine of negative 1/6. I encourage you to verify this on your own; on a calculator, you would get that that's approximately 0.22.

Then, pi over 4 is approximately 0.785, so this expression would be approximately 0.22 minus 0.7 times n, where n could be any integer. Then, this one over here on the right, let me do that in the yellow, x could be approximately equal to... well, if this evaluates to approximately 0.22, then this is just the negative of it, so it's going to be negative 0.22.

Then, it's plus what approximately pi over 4 is, so 0.785n. Now, what we could do is just try different n's and see if we're starting above or below this interval, and then see which of the x values actually fall in this interval. So let's just start here. If we just start at n equals zero... actually, when I set up a little table here, if we have n here and if we have the x value here.

When n is zero, well then you don't see this term, and you just get approximately 0.22. Now let's compare that to the interval; the upper bound of that interval is zero, so this does not sit in the interval. This is too high, and we would want to define the x's that sit in the interval. We want to find lower values, so it's good that here we're subtracting 0.785.

I would use positive integer values of n to decrease this 0.22 here. So when n equals 1, we would subtract 0.785 from that, and I'll round all of these to the hundredths place. That would get us to negative 0.57, and that does sit in the interval, so this looks good. This would be a solution in that interval right over here.

Let's try n equals 2, so we would subtract 0.785 again, and that would get us to negative 1.35, not 2.5, 3.5, and that also sits in the interval. It's larger than negative 1.57, so that looks good. Let's subtract 0.785 again when n equals 3; that would get us to negative 2.14. Well, that's all of a sudden out of the interval because that's below the lower bound here, so this is too low.

So we've been able to find two x values that sit in the interval that we cared about. Now, let's use these x values right over here, and I'll set up another table. So let's see, we have our n and then we have our x values. So let's start with n equals 0 because that's easy to compute. Then this term would go away, and we'd have negative 0.22.

That's actually in this interval here; it's below zero; it's larger than negative 1.57, so that one checks out. But now, to really explore, we have to go in both directions. We have to increase it or decrease it. So if we wanted to increase it, we could have a situation where n equals 1.

So if n equals 1, we're going to add 0.785 to this. Now, you immediately know that that's going to be a positive value. If you computed it, it'd be 0.57, which is larger than zero, so this is too high. So now we could try going lower than negative 0.22 by having negative values of n.

So if n is equal to negative 1, that means we're subtracting 0.785 from this right over here, which would get us to negative 1.01. Well, that one works out, so that's in our interval. Now let's subtract 0.785 again, so I'll have n equals negative 2. If I subtract 0.785 again, I could round that to negative 1.79, which is lower than negative 1.57, so it's out of our interval, so it's too low.

All of the x values that are in our interval that satisfy this equation are these two right over here and this one, and this one. And we are done.