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## Precalculus

### 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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# Sine equation algebraic solution set

CCSS.Math:

Solve a sine 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?

- In the exercises, +2πn (or +360n) is added directly to the right side of the equation, but in this video, it is added first to x inside the brackets then later on transposed to the opposite side so it became -2πn (or -360n).

I got the same numbers but the only difference is the sign. For example, in one of the exercises, I got the solution: -6.07-45n following Sal's method. But in the choices, it only provides -6.07+45n. When I checked the explanation, the +360n is added directly on the right side. Can someone explain which way is the correct evaluation of the solution?(10 votes)- Since 𝑛 can be any integer, positive or negative, it doesn't matter whether we write +45𝑛 or −45𝑛.

I mean, 45𝑛 = −45⋅(−𝑛), and if 𝑛 is an integer, then −𝑛 is also an integer.(5 votes)

- How and when do we differentiate between adding the 2pi(n) to the X side or the opposite side.

In previous lessons the 2pi(n) was added to the right side.

When I did the problem before watching this lesson I ended up with x= 4sin^-1(3/8+2pi(n)) and x=-4sin^-1(3/8-2pi(n)) as I thought you just add that bit in there but now I am getting confused as when to know exactly how and when to add this step.. (the 2pi(n) and such)(4 votes)- When solving trigonometric equations, it is important to remember that the trigonometric functions are periodic, meaning they repeat themselves after a certain interval. For example, the sine function has a period of 2π, which means that sin(x) = sin(x + 2π) for any value of x.

When solving trigonometric equations, we may need to find all solutions that satisfy the equation. To do this, we often use the general solution, which includes all possible solutions, and then use the periodicity of the trigonometric functions to find all solutions in a given interval.

When using the general solution, we add 2πn (or a multiple of the period) to the angle on the right-hand side of the equation. This is because adding 2πn to the angle will give us another angle that has the same trigonometric function value.

For example, if we have the equation sin(x) = 0.5, we can use the inverse sine function to find one solution: x = sin^-1(0.5) = π/6. But since the sine function has a period of 2π, we know that there are other angles that have the same sine value, such as x = 5π/6, 13π/6, etc. To find all solutions, we use the general solution:`x = π/6 + 2πn or x = 5π/6 + 2πn`

where n is an integer. This gives us all possible solutions to the equation.

So, to answer your question, we add the 2πn to the angle on the right-hand side of the equation when using the general solution to find all solutions to a trigonometric equation. It is important to keep track of which quadrant each solution is in, as this will affect the sign of the trigonometric function.(3 votes)

- In the exercises, it asked me to find
*all*solutions to a sine equation like the one in this video and I did get the sin(θ+2πn) answer correct but I couldn't find the sin(π-θ+2πn) answer in the provided answers, and under the hints it says that we only find the answers in the -π ≤ θ ≤ π interval, and that the first answer was the "trough" and the only correct one. I would appreciate any help that explains this.(3 votes)- If the first solution was either 𝜃 = 𝜋∕2 or 𝜃 = −𝜋∕2, then there is no integer 𝑛 for which 𝜋 − 𝜃 + 2𝜋𝑛 is a different solution in the interval [−𝜋, 𝜋].

In mathematics a trough is basically the bottom of a curve, in this case the bottom of the unit circle, i.e. 𝜃 = −𝜋∕2(3 votes)

- At3:59, Sal takes the
**inverse sine**of both sides of the equation. He assumes that arcsin(sin(...))=... on the left side.

However, as mentioned in the previous videos, the**range**of the arcsine function is -π/2 to π/2. This means that if you take arcsin(sin(θ)), you don't always get the result θ.

I understand Sal intends to get the complete solution set, but I wonder if he is being**mathematically rigorous**at this.(3 votes)- Yes, but in since this video is about a solution set, the specific value of sine(theta) is irrelevant, but the set of all the values is.(2 votes)

- I dont get why on the exercises they say we use sin(180-O) but then in the process they use sin(-180 -O)(4 votes)
- sin(180-theta) = sin(-180-theta)

Think about where the angle -180/180 degrees is on the unit circle(0 votes)

- 4:08, Sal takes the inverse sine of both sides of the equation, but the inverse sine limit the value, in this case the 3/8, to the first and fourth quadrants. Will doing so have any effect on whether the solutions we are valid?(2 votes)
- That is why n and other solution involving sin(180-theta) is there.(1 vote)

- shouldn't there be 4 equations, 2 for positive sin value and 2 and negative sin values?(1 vote)
- sin(𝑥∕4) = 3∕8 tells us that the sine value is positive.

Hence, the angle 𝑥∕4 is either in Quadrant I or Quadrant II.(2 votes)

- At3:09We can get sin(-pi-x/4+2pi*n) = 3/8,instead sin(pi-x/4+2pi*n) = 3/8 ??(1 vote)
- Hi, yes that is correct you can do -pi instead pi.(1 vote)

## Video transcript

- [Instructor] The goal of this video is to find the solution set
for the following equation. So all of the X values, and we're dealing with radians, that will
satisfy this equation. So I encourage you, like
always, pause this video and see if you can work
through this on your own before we worked through it together. All right, now let's
work through it together. Now your intuition, which
would be correct, might be, let's see if we can isolate the sin of X over four algebraically. And the first step I
would do is subtract 11 from both sides. And if you do that, you
would get eight sin of X over four is equal to three,
just subtracted 11, both sides. Now to isolate the sin, I would
divide both sides by eight and I would get sin of X
over four is equal to 3/8. Now, before I go further,
let's think about whether this is the most
general solution here or whether we're going to find
all of the solution set here. Well, we have to remind ourselves, let me actually draw a little
bit of a unit circle here. So that's my X axis. That's my Y axis. Then a circle. And if we have some angle
theta right over here. So that's theta, we know that
the sin of theta is equal to the Y coordinate of where this radius
intersects the unit circle. And we also know that if we add an arbitrary
number of two pies here or if we subtract an
arbitrary number of two pis, we go all the way around the unit circle, back to where we began and so the sin of theta would be the same. So we know that sin of theta
plus any integer multiple of two pi, that's going to
be equal to the sin of theta. And so we can generalize
this a little bit. We can instead of just
saying, sin of X over four is equal to 3/8, we could
write that sin of X over four plus any integer multiple of
two pi is going to be equal to 3/8 where N is any integer. It could even be a negative
one, a negative two or of course it could be
zero, one, two, three, so on and so forth. So is this, if we solve now for X, is this going to give us the
most general solution set? Well, we can also remind ourselves
that if I have theta here and sin if theta gets there. There's one other point on the unit circle where I get the same sin. It would be right over here. The Y coordinate would be the same. And one way to think about it is if we start at pi radians,
which would be right over there and we were to subtract theta, we're going to get the same thing. So this angle right over here, you could view as pi minus theta. And you could keep trying
it out for any theta, even the theta that put
you in the second quadrant, third quadrant or fourth quadrant,
if you do pi minus theta, sin of pi minus theta, you're going to get the same sin value. So we also know that sin of pi minus theta is equal to sin of theta. And so let me write another
expression over here. So it's not just sin of X over four is equal to 3/8. We could also write that
sin of pi minus X over four, 'cause X over four is the theta here. Sin of pi minus X over
four is equal to 3/8. And of course we can also
use the other principle that we can add 2 pi or
subtract two pi from this, an arbitrary number of times and the sin of that will
still be equal to 3/8. So I could write it like this. Sin of pi minus X over four
plus an integer multiple of two pi, that is going
to be equal to 3/8. And if I solve both of these, the combination of them, the union of them would give me the broadest solution set. So let's do that. So over here, let me take the
inverse sin of both sides. I get X over four plus two pi N is equal to the inverse sign of 3/8. Now I could subtract two
pi N from both sides. I get X over four is equal to the inverse sin of 3/8, minus two pi N. And if you think about it,
because N can be any integer, this sin here in front of the two, this negative really doesn't matter. It could even be a positive. And now let's multiply both sides by four. We get X is equal to four
times the inverse sin of 3/8 minus eight pi N. And then if I work on this
blue part right over here, same idea, take the
inverse side of both sides. We get pi minus X over
four plus two pi N is equal to the inverse sin of 3/8. And then let's see, I can
subtract pi from both sides and subtract two pi N from both sides. And so I get negative X over four is equal to the inverse sin of 3/8
minus pi minus two pi N. And I multiply both
sides by negative four. I get X is equal to negative
four times the inverse sin of 3/8 plus four pi plus eight pi N. And as I mentioned, the
union of both of these give us the entire solution set to our original equation here.