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### Course: Precalculus > Unit 2

Lesson 7: Sinusoidal equations- Solving sinusoidal equations of the form sin(x)=d
- Cosine equation algebraic solution set
- Cosine equation solution set in an interval
- Sine equation algebraic solution set
- Solving cos(θ)=1 and cos(θ)=-1
- Solve sinusoidal equations (basic)
- Solve sinusoidal equations

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# 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.

## Want to join the conversation?

- How does Sal get 1/8 and -1/8? Does that mean that there is an invisible 1 before the cosine?(5 votes)
- There is always an invisible 1, because 1 times anything = the same as before, so I can multiply anything by 1.

However, this is not the reason he used 1/8. Simply put, multiplying the rest of the equation by 1/8 is the same as dividing by 8. The answer will be the same.

Sal probably wrote it as a fraction so it looks a bit neater.

This is very basic math stuff I'm talking about, maybe 4-5th grade.

I highly recommend you revisit those basic math essentials before doing Trigonometry, or you will be confused and stuck on a lot of things.(10 votes)

- 8cos(12x)+4=−4 at 1 point the explanation of the mastery question says

Since −1-1−1minus, 1 is a trough, it is the only solution within this interval.

What is meant by trough? Does it indicate the lowest value in the wave graph? What about the equation would tell me that? (assuming that is the correct interpretation of the word 'trough').(10 votes)- It would be helpful to learn the sinusoidal expression's anatomy before doing this lesson.

I assume a "trough" would be the minimum part of the graph, which can be calculated by subtracting the absolute value of the amplitude from the midline.

For example, in the function 3 cos(x)+5, 3 is the amplitude and 5 is the midline.

Without even graphing this function, it is clear that the maximum value is 5 + 3, or 8, and the minimum value as 5 - 3, or 2.

You can graph this function and you'll see I'm correct.

I am not entirely sure what the question you had was, but hopefully this helped you (assuming you're still stuck 2 years later lol)(2 votes)

- Where did the -(𝜋/4)n and +(𝜋/4)n come from?(6 votes)
- When Sal divides both sides by 8 and -8 the omitted step will be (I'll write for 8 but it's applies to -8 as well): (cos^-1(-1/6)-2pi * n)/8. So if you remember properties of ratios(sorry don't remember correct term) it'll could be also written as: (cos^-1(-1/6))/8 - (2pi * n)/8. In the second term(subtrahend) both 2(in numerator) and 8(in denominator) could be divided by 2 leaving us 1 and 4 respectively, which leads us to (1pi *n)/4 and since 1 is often omitted as a coefficient we have (pi * n)/4.(4 votes)

- why negative cos theta is equal to positive cos theta?(3 votes)
- −cos(𝜃) ≠ cos(𝜃)

But cos(−𝜃) = cos(𝜃)

Draw 𝜃 and −𝜃 in the unit circle, and you'll see that they have the same 𝑥-coordinate.(7 votes)

- It's incorrect to apply arccos(cos(θ))=θ to cos(-8x+2pi*n), as "-8x+2pi*n" represents angle which is outside of range of arccos function. "-8x+2pi*n" belongs to the third quarter of the unit circle, while arccos is able to spit out only angles that belong to the first and the second quarter of the unit circle.(4 votes)
- It's the other way around.

−8𝑥 could very well be outside the arccos range, which is why we can't say

arccos(cos(−8𝑥)) = −8𝑥

But for any given 𝑥 there will always be exactly one integer 𝑛, such that

arccos(cos(−8𝑥 + 2𝜋𝑛)) = −8𝑥 + 2𝜋𝑛(3 votes)

- if my answer are x=1/8cos^-1(-1/6) + pi/4n and x=-1/8cos^-1(-1/6) + pi/4n also correct ? because if you add or minus 2pi it also the same ? i normally add instead of minus(4 votes)
- i belive that is correct.(2 votes)

- In one of the practices, there is a question I am confused about.

The question was 6sin(3x)+1=7 (In degrees)

I got the solution sets x = 30 * 360n & x = (180 - 30) * 360n

The correct answer out of the choices was x = 30 + 120n. However, when you take the sin(30 + 120n) and have n take on the values 2 + 3x (x being an integer =< 0) you get -1, instead of 0.5, what you should get.

Can someone please tell me what I am not understanding?(2 votes)- n take on the values 2 + 3x.So this would mean

x = (n-2)/3 which does not make sense. We want x to be the subject not n.

You need let n be an integer.

So the correct answer is x = 30 + 120n.

n = 1

=> x = 150

6*sin(450)+1 = 7 since sin(450)=sin(90)=1

I have various answers on sinusoidal equations in my profile that might be useful to look at.(2 votes)

- why is the answer set in the 2nd and 3rd quadrant when the cos function is limited to the 1st and 2nd?(2 votes)
- Suppose we have cos(theta) = -1/6. Then any solution will in the 2nd and third quadrant. This comes from the unit circle definition, cos(theta) = x. Since x is only negative in 2nd/3rd quadrant all solutions are only in those two quadrants.(2 votes)

- Hi, I was just wondering why we had to include the x values that satisfy the equation cos(-8x)= -1/6 in our final answer. I do understand that cosine is an even function, therefore cos(x) = cos(-x), but aren't we only searching for the x values that satisfy cos(8x)=-1/6? timestamp at2:40approximately(2 votes)

## 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.