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# Solving sinusoidal equations of the form sin(x)=d

## Video transcript

which of these are contained in the set which of these are contained in the solution set to sine of X is equal to one-third and answers should be rounded to the nearest hundred select all that apply and so I encourage you to pause the video right now and work on it on your own so I'm assuming you've given a go at it let's think about what this is asking so they're asking what are the x values what is the solution set what are the possible X values where sine of X is equal to one-third and to help us visualize this let's draw a unit circle so that's my y-axis this right over here is my x-axis this is X at 1 this is y positive 1 negative 1 along the x axis negative 1 on the y axis a unit circle I'm going to Center it at 0 it's going to have a radius of 1 radius of 1 and we just have to remind ourselves what the unit circle definition of the sine function is if we have some angle we have some angle one side of the angle is going to be array along the positive x axis we do this in a color you can see along the positive x axis and then the other side so let's see this is our angle right over here let's say that's some angle theta the sine of this angle is going to be the y-value of where this ray intersects the unit circle so this right over here that is going to be sine of theta so with that review out of the way let's think about what X values and we're assuming we're dealing in radians what X values what if I take the sine of it is going to are going to give me 1/3 so when does y equal 1/3 along the unit circle so that's 2/3 1/3 right over the year we see it equals 1/3 exactly two places here and here so there's two angles where or at least two if we just take one or two on each pass of the unit circle then we can keep adding multiples of 2/pi to get as many as we want but we see just on the unit circle we could have this angle we could have this angle right over here or we could go all the way around to that angle right over there and then we could add any multiple of 2 pi to those angles to get other angles that would also work where if I took the sine of them I would get 1/3 now let's think about what these are and here we can take our calculator out and we could take the inverse sine of 1/3 so let's do that the inverse sine of 1 over 3 and we have to remember what the range of the inverse sine function is it's going to give us a value between negative PI over 2 and PI over 2 so a value that sticks us and either the first or the fourth quadrant if we're thinking about the unit circle right over here and so we see that gave us zero point if we round to the nearest hundredth 3 4 so essentially they've given us this value they've given us zero point 3 4 that's this angle right over here how did I know that well it's a positive value it's greater than zero but it's less than PI over 2 pi is 3.14 so PI over 2 is going to be one point one point five seven and we can go on and on and on so this right over here is zero point three four radians but what would this thing over here be well it's going to be whatever if we go to the positive if we go to the negative x-axis and we subtract zero point three four so we subtract a zero point two three four so this is zero point three four we're going to get to this angle so it's going to be if we take pi minus the previous answer it gets us free rather than ears hundreds it's two point eight radians so this is zero point three four radians and then this one let me do it in this purple color this one right over here if we were to go all the way around it's pi minus zero point three four which is two point eight zero radians rounding to the nearest hundredth now that's not all of the values we can add multiples of 2/pi to each of these so two point eight zero plus any multiple of two pi so two pi n where n is an integer n is an integer or we could take zero point three four and add any multiple of two pi so two pi n where n is N integer so our solution set here just to write it rewrite it in a kind of outside of this messiness it's going to be two point eight zero radians plus 2pi and where n is an integer and zero point three four plus two pi n where n is an integer so let's see which of these are at least a subset of this so we look at 0.34 plus two pi n where n is an integer well that's exactly what we wrote over here so that's zero point three four if N is a positive integer we'll go around this way and we keep getting back to the same point if it's a negative integer we go around that way we keep getting to the same point but that's definitely in the solution set zero point three four plus PI n four and an integer so if we have zero point three four and if we were to not add two pi but just pi where would we get two well we would get two right over there and the sign of this isn't going to be positive one third it's going to be negative one third so we could rule that out negative zero point three four well that's this angle right over here the sine of that's going to be negative one third and if you add a multiple of 2pi to that you're still going to get negative one third so that doesn't work same thing for this one right over here two point eight plus two pi n that's what we wrote right over here two point going all the way to two point eight and any multiple of it is going to get you back to that same point so that one works two point eight plus PI n so if you're here if you added PI you're going to get over here and the sine of that isn't going to be positive one third it's going to be negative one third so we can rule this one out as well so these are the only two apply and if you actually take them together you have the entire solution set to this equation right over here