Sine equation algebraic solution set | Trigonometry | Precalculus | Khan Academy

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 work 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 sine of x over 4 algebraically. The first step I would do is subtract 11 from both sides. And if you do that, you would get 8 sine of x over 4 is equal to 3. Just subtracted 11 both sides. Now, to isolate the sine, I would divide both sides by 8, and I would get sine of x over 4 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, a circle, and if we have some angle theta right over here, so that's theta, we know that the sine 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 2 pi's or if we subtract an arbitrary number of 2 pi's, we go all the way around the unit circle back to where we began, and so the sine of theta would be the same. So we know that sine of theta plus any integer multiple of 2 pi, that's going to be equal to the sine of theta.

And so we can generalize this a little bit. We could, instead of just saying sine of x over 4 is equal to 3/8, we could write that sine of x over 4 plus any integer multiple of 2 pi is going to be equal to 3/8, where n is any integer. It could even be a negative 1, a negative 2, or, of course, could be 0, 1, 2, 3, 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 sine of theta gets there, there's one other point on the unit circle where I get the same sine. It would be right over here; the y-coordinate would be the same.

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. 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, sine of pi minus theta, you're going to get the same sine value.

So we also know that sine of pi minus theta is equal to sine of theta. And so let me write another expression over here. So it's not just sine of x over 4 is equal to 3/8. We could also write that sine of pi minus x over 4, because x over 4 is the theta here, sine of pi minus x over 4 is equal to 3/8.

And of course, we can also use the other principle that we could add 2 pi or subtract 2 pi from this an arbitrary number of times and the sine of that will still be equal to 3/8. So I could write it like this: sine of pi minus x over 4 plus an integer multiple of 2 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 sine of both sides. I get x over 4 plus 2 pi n is equal to the inverse sine of 3/8.

Now I could subtract 2 pi n from both sides. I get x over 4 is equal to the inverse sine of 3/8 minus 2 pi n. And if you think about it, because n can be any integer, this sign here in front of the 2, this negative really doesn't matter. It could even be a positive, and now let's multiply both sides by 4.

We get x is equal to 4 times the inverse sine of 3/8 minus 8 pi n. And then, if I work on this blue part right over here, same idea: take the inverse sine of both sides. We get pi minus x over 4 plus 2 pi n is equal to the inverse sine of 3/8.

And then let's see, I could subtract pi from both sides and subtract 2 pi n from both sides, so I get negative x over 4 is equal to the inverse sine of 3/8 minus pi minus 2 pi n. I multiply both sides by negative 4; I get x is equal to negative 4 times the inverse sine of 3/8 plus 4 pi plus 8 pi n.

And as I mentioned, the union of both of these give us the entire solution set to our original equation here.