Cosine equation algebraic solution set

The goal of this video is to find the solution set for the following equation: negative 6 times the cosine of 8x plus 4 is equal to 5. And like always, I encourage you to pause this video and see if you can have a go at this before we do it together. A reminder: we want the entire solution set, not just one solution.

All right, now let's work through this together. Some of you might recognize that it would be valuable to isolate the cosine of 8x, and a good way of doing that would be first to subtract 4 from both sides. That would get us negative 6 times cosine of 8x. I subtracted 4 from the left, so that 4 is going to be gone, and then if I subtract 4 from the 5, I'm going to get a 1 there.

Now I can multiply both sides of this equation by negative 1. 6. I just want to have a 1 in front of the cosine, so negative 1. 6. This is going to be 1, so I'm just going to have cosine of 8x is equal to negative 1/6. Now, if I just keep going, I could take the inverse cosine of negative 1/6. Whatever that is, divided by 8, I would get a solution, but this is a good time to pause and to make sure that we're capturing all of the solutions.

I'll refresh our memories with some identities. To help with these identities, I like to draw a quick unit circle. So this is our x-axis, this is our y-axis, and so my quick hand-drawn unit circle might look something like this. It's not that nice looking, but we want to think about all of the angles that when I take the cosine, I get to negative 1/6.

So negative 1/6 might be something like right over here. You can see that there might be an angle like this that would get us there. So let me draw that; draw the radius. We know the cosine of an angle is the x-coordinate of where that radius, defined by that angle, intersects the unit circle. But we also see there's another place: if we essentially take the negative of that angle, we could go right over here, and we would also get the same cosine.

So we could go the negative of the angle; go that way, and that's where we get the identity that cosine of negative theta is equal to cosine of theta. Hence, if cosine of 8x is equal to negative 1/6, using this identity, we also know that cosine of the negative of this will also be equal to negative 1/6. So let me write that down: cosine of negative 8x is also going to be equal to negative 1/6.

Now already, we have expanded our solution set because this is going to give us another x value that's going to get us the result that we want. But are we done? Well, the other thing to realize is, let's say I have some angle here. If I take the cosine, I get to negative 1/6, but then if I add 2π, again I'm going to get to the same place, and the cosine is once again going to be negative 1/6.

I could add 2π again; I could essentially add 2π an arbitrary integer number of times. So I could rewrite this right over here as cosine instead of just 8x; it's 8x plus an integer multiple of 2π. That's also going to be equal to negative 1/6. Similarly, for negative 8x, I could say cosine of negative 8x plus an integer multiple of 2πn is going to be some integer in both of these situations that’s also going to get us to negative 1/6.

Now we can feel pretty good that we're capturing all of the solutions when we solve for x. So in both of these, let's take the inverse cosine of negative 1/6 in order to solve for x here. If we were to take the inverse cosine of both sides, we could get that x plus 2π times some arbitrary integer n is equal to the inverse cosine of negative 1/6. Now let's solve for x: we can subtract 2πn from both sides, so we could get 8x is equal to the inverse cosine of negative 1/6 minus 2πn.

Now, it's interesting to note that the sign on this 2πn term actually doesn't matter so much because n could be a negative integer, but I'll just stick with this negative 2πn. If we wanted to solve for x, we just divide both sides by 8; we get x is equal to 1/8 times the inverse cosine of negative 1/6 minus π/4n.

Now we can do the exact same thing in the other scenario. I'll call this the yellow scenario, where if I take the inverse cosine, I get negative 8x plus 2πn is equal to the inverse cosine of negative 1/6. Now I can subtract 2πn from both sides, so I get negative 8x is equal to inverse cosine of negative 1/6 minus 2πn.

Now I can multiply both sides by negative 1/8 or divide both sides by negative 8, and I get x is equal to negative 1/8 times the inverse cosine of negative 1/6 plus π/4n. I will stop here for this video. At least algebraically, we know the solution set, and this is the complete solution set. If you take the combination of both of these x values, if you take the combination of both of these expressions, in a future video, we'll evaluate this with a calculator, and we'll think about the solutions that fit within a given interval.