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### Course: Trigonometry > Unit 4

Lesson 2: 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 solution set in an interval

Given the algebraic solution set for a cosine equation, find which solutions fall within an interval. Created by Sal Khan.

## Want to join the conversation?

- The interval (-pie/2,0),is arbitrary?. Could he have chosen any other interval?(16 votes)
- other people have asked the same problem check their answers and hopefully that will help you. I am asking myself the same question so I dont trust myself to answer(7 votes)

- 1:46, Sal encourages us to evaluate a portion of the equation on a calculator. Instead of coming up with approximately 0.22, I keep ending up at approximately 12.44. Can anyone explain why?(9 votes)
- 0.22 radians = 12.44°

Change the calculator mode to "RAD" instead of "DEG".(21 votes)

- I really don't understand why he choose that interval, what was the purpose of choosing it? since from the previous video he already graphed it a little past pi/2, but here he chose a invertval that does not even reach it?(14 votes)
- Hi, why are we restricting the range of x to be in between [-pi/2, 0]? timestamp at approximately0:31(9 votes)
- i think he chose arbitrary values for the range. the point of this video is just to teach us how to find values of x within a range(8 votes)

- Where did the interval [-π/2, 0] come from?(7 votes)
- That was just part of the given of the problem which puts our cos in the 3rd and 4th quadrant. So cos-1(x) will have positive values in the 1st and 4th quadrants and negative values in the 2nd and 3rd quadrants. Sal uses this limit to say we are only thinking about what is happening in the 3rd and 4th quadrant.(5 votes)

- @0:52, why did he take -pi/2, was it just an arbitrary interval?(9 votes)
- Hi, why did SAL take [-Pi/2, 0] as the limit. Shouldn't the limit for the COS function be between 0 and Pi (0 and 3.14) ?

Considering the given value of -1/6 the coordinate turns out to be falling on second quadrant. But the same coordinate can be achieved on the vertically opposite side that is third quadrant. Since both these quadrants (second and third) give the stated value in the question (i.e. -1/6), I was assuming we should have restricted to one of these quadrant. Can someone help clarify this?(6 votes)- The COS function (adj/hyp) on the unit circle is positive in the 1st and 4th quadrants, and negative in the 2nd and 3rd quadrants. If you went between the 1st and 2nd, the cos(first)=-cos(second), so you get two different answers.(3 votes)

- Why can't there be infinite x values that satisfy these situations? For example, if 0 & 1 satisfy it; why can't 0.67 satisfy it and every other decimal quantity in between 0 and 1?(5 votes)
- That's not how functions work. If two values satisfy a condition, everything in between needn't.

Think about this. Say I have x^2 = 4. Both x = -2 and x = 2 satisfy this. However, no number between -2 and 2 satisfies it.(5 votes)

- hi, can you please tell me weather (negative pi over 2 and zero) is the domain or the range?(4 votes)
- In this case it is neither. We only consider domain and range when we are considering functions. In this case however we are focused on finding a solution in a particular interval.(4 votes)

- I plugged in 1/8inversecos(-1/6) into my calculator but I got 12.44925853...but he's saying it's 0.22. Can someone explain this?(2 votes)
- yeah, like rhlliu100 said, u have to use radians. i need to be very careful about that cuz im gonna take ap calc bc exam, and i always find my calc on degrees when it should be in radians :)(7 votes)

## Video transcript

- [Instructor] In a previous
video, we established the entire solution set
for the following equation. And 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 as numerous xs is that for
any integer value of n, you'll get another solution. For any integer value of n,
you'll get another solution. What I wanna do in this video is to make things a
little bit more concrete. And 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 two to zero. 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. So 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 two, 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 -1.57 to zero. - 1.57 isn't exactly negative pi over two, but it'll hopefully be suitable for what we're trying to do here. And now let's see if we can write the different parts of these expressions, or at least approximate them as decimals. So this could be rewritten
as x is approximately, if you were to take 1/8
times the inverse cosine of -1/6, I encourage you to verify this on your own on a calculator, you would get that that's
approximately 0.22. And then pi over four
is approximately 0.785. So this expression would
be approximately 0.22 minus 0.785 times n, where n could be any integer. And then this one over here on the right, let me do that in this 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 -0.22. And then it's plus what approximately pi over four is, so 0.785n. And 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 why don't I set
up a little table here, 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. So this is too high and we would want to define the x's that sit in the interval. We wanna find lower values. So it's good that here, where
you're subtracting 0.785, so I would use positive
integer values of n to decrease this 0.22 here. So when n equals one, we would
subtract 0.785 from that, and I'll round all of these
to the hundredths place, and that would get us to -0.57, and that does sit in the interval. So this looks good. So this would be a solution in that interval right over here. And let's try n equals two. So we would subtract 0.785 again, and that would get us to -1.35, not 25, 35, and that also sits in the interval. It's larger than -1.57,
so that looks good. Let's subtract 0.785
again, when n equals three, that would get us -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 using this expression,
we've been able to find two x values that sit in the
interval that we care 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
zero 'cause that's easy to compute, and then
this term would go away, and we'd have -0.22, and that's actually in this interval here, it's below zero, it's larger than -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 one. So if n equals one, we're
gonna 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 -0.22 by having negative values of n. So if n is equal to -1, that
means we're subtracting 0.785 from this right over here
which would get us to -1.01. Well, that one works out,
so that's in our interval. And now let's subtract 0.785 again. So I'll have n equals -2. And so if I subtract 0.785 again, I could round that to -1.79, which is lower than -1.57, so it's out of our
interval, so it's too low. So 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.