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## Precalculus

### Course: Precalculus>Unit 2

Lesson 7: Sinusoidal equations

# Cosine equation algebraic solution set

Solve a cosine equation with an infinite number of solutions. Use trig identities to represent the whole solution set. Created by Sal Khan.

## Video transcript

- [Lecturer] The goal of this video is to find the solution set for the following equation, negative six times the cosine of 8x plus four is equal to five. 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. And 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 four from both sides, and then that would get us negative six times cosine of 8x, I subtracted four from the left, so that four is going to be gone, and then if I subtract four from the five, I am going to get a one there. And now I can multiply both sides of this equation by negative 1/6, I just wanna have a one in front of the cosine, so negative 1/6. And so this is going to be one, so I'm just gonna 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, and whatever that is divided by eight, I would get a solution, but this is a good time to pause and to make sure that we are capturing all of the solutions. And I'll give us, or I'll refresh our memories with some identities. And 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, (laughing) it's not that nice looking, but we wanna 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. And so 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 that's defined by that angle, where that radius 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 to 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. And so 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, where if I take the cosine, I get to negative 1/6, but then if I had two pi again, I'm gonna get to the same place, and the cosine is, once again, going to be negative 1/6, and I could add two pi again, I could essentially add two pi 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 two pi, that's also going to be equal to negative 1/6. And similarly for negative 8x, I could say cosine of negative 8x plus an integer multiple of two pi, and is going to be some integer in both of these situations, that's also going to get us to negative 1/6. And so now we can feel pretty good that we're capturing all of the solutions when we solve for x. So in both of these, now let's take the inverse cosine of negative 1/6 in order to solve for x here. So if we were to take the inverse cosine of both sides, we could get that 8x plus two pi times some arbitrary integer n is equal to the inverse cosine of negative 1/6. And then now let's solve for x, we can subtract two pi n from both sides. So we could get 8x is equal to the inverse cosine of negative 1/6 minus two pi n. Now, it's interesting to note that the sign on this two pi n term actually doesn't matter so much, 'cause n could be a negative integer, but I'll just stick with this negative two pi n. And so if we wanted to solve for x, we'd just divide both sides by eight, we get x is equal to 1/8 times the inverse cosine of negative 1/6 minus pi over four n. And 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 two pi, n is equal to the inverse cosine of negative 1/6. And now I can subtract two pi n from both sides, so I get negative 8x is equal to inverse cosine of negative 1/6 minus two pi n. Now I can multiply both sides by negative 1/8, or divide both sides by negative eight, and I get x is equal to negative 1/8 times the inverse cosine of negative 1/6 plus pi over four n. So I will stop here for this video, where at least algebraically we know the solution set, and this is the complete solution set 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.