>> Solving trigonometric equations is precisely the same in concept as solving equations from the past. That is, we have some variable involved and we're looking for the value or values of the variable that will cause the statement of equality to be true. Consider this problem. Now, notice that the variable involves sine. So we have sine of some variable and we have some other stuff here. Well, we'd like to isolate the sine of X before we evaluate the value for X and it's pretty easy to evaluate X once we have done that. Adding 1 on both sides we get 2 sine X equals 1. Dividing on both sides by 2 to clear this coefficient, we find that sine X is equal to 1/2. Now, so the sine of what angle or the sine of what real number is equal to 1/2 is what this is requesting. Now, let's find an exact value and let's evaluate this using our little triangles for reference and our unit circle, and we could use a graphing approach to analyze the situation. And all of those ideas will come in handy in this situation. Well the first thing we might think about is that well, let's see. The sine of some angle is 1/2 and I know that that angle is found in some little triangle having to do with this 1, 2 square root of 3 business and the 30-degree angle is here and from the perspective of that 30-degree angle, the opposite side over the hypotenuse is 1/2. So, aha, it's a 30-degree angle here that I'm talking about for X. Or it's an angle of pi over 6. So, X is either 30 degrees or pi over 6. Since it's a letter that's being used here and not a theta symbol, I'm gonna assume that it's pi over 6 that we're looking for. So, I'll say X is pi over 6. Now, we want all the values of X that will make this true and we know that if we examine the unit circle situation that sine is positive in this quadrant and in this quadrant but negative in these two. So, maybe there's an angle in the second quadrant that also makes this true and certainly there is. You see we have found the angle in the first quadrant. This 30-degree angle makes that true, but there is a reference angle of 30 degrees in the second quadrant that also makes that true. Now, reference angle of 30 degrees but what's this angle? That's the angle that we want, because it's the sine of this angle that will also be 1/2. Well, let's see. If we're talking 30 degrees then this angle would be 180 minus 30 or 150 degrees. If we're talking about pi over 6 for the reference angle and we know it's pi up to here then pi or it's 6 pi over 6 minus 1 pi over 6 is 5 pi over 6. So, an angle of 5 pi over 6 would also make this true. Are there any others? Well, sure. Every angle which is co-terminal with these two will also make this statement of equality true. That is if we are right here, then--then an angle which goes all the way around and back to that spot, you see, will make it true and go around twice to the spot and it'll make it true and so on and the same idea for this. So we can generate now the infinitude of answers, the infinite idea that we have a whole bunch of values of X associated with pi over 6 and a whole bunch associated with 5 pi over 6 that make this statement of equality true. So let's generate those. It would be pi over 6 plus--Now let me emphasize. I wanna make sure we understand where I'm going with this conversation. That it is pi over 6 and then, pi over 6 plus 2 pi, you see, go all the way around the circuit and then it's pi over 6 plus 4 pi. You see, and then pi over 6 plus 6 pi and so on. So we wanna write some statement that gives all of these values in one little statement. And to do that, I think we've seen the notation before but it's--it's pi over 6 plus N times 2 pi. You see N being an integer. Now, that integer, you know, can be positive or negative. So we can generate values in either direction, you see, and this gives the--the infinite answers associated with pi over 6. And with 5 pi over 6, same idea. Now a lot of times this part of the problem is--is rearranged, these factors are rearranged and its 2 pi N or 2N pi, something like that. So we could write this as, in another way, as 2 pi N if we wanted to do that.