>> This next equation is quite obviously in quadratic form. We have a 2nd degree term involving sign X. We have a 1st degree term involving sign X, and we have a constant term. And we're gonna use the same techniques that we would use in the past for solving quadratic equations. Our first choice would be to solve by factoring, and if we absolutely have to, if we can't factor, we have to use the quadratic formula. Now do avoid using the quadratic formula unless you absolutely have to here, because it just creates some approximations that you don't want to deal with unless you absolutely have to deal with them. At any rate, this one turns out to be factorable. If we can't see through the factorization very easily, then we just replace sign X with a letter like N. We might think of this as 2N squared plus 3N, then plus 1 equals 0, and factor this. And then replace N with sign X, and we have the factored form in terms of what we have here. But I think we can see through this, or quite possibly we can. The factored form turns out to be, well let's see... I'm thinking let's see, for 2 signs squared X it's 2 sign X times sign X. Like 2N times N for 2N squared, same idea. Alright, all signs are plus and we have 1 and 1 for the factors of the last term. And we can verify the factorization by the multiplying and getting the middle term. Okay, now the idea is that we have solutions at the 0's of each factor, so we'll take each factor and set them equal to 0. [ Writing on the board ] And let's see, for this one, if we're solving we'll subtract 1 on both sides. [ Writing on the board ] And then dividing on both sides by 2... Now let's see, let's think about sign X is negative... in what quadrant? Well let's see where, and what kind of angle are we talking about here for this angle X? Well let's see, for one-half we're thinking maybe this way... It's, remember the 1, 2, square root of 3, and the idea that sign is opposite over hypotenuse. So from the perspective of this angle, we have a one-half. Ah, 30 degrees or pi over 6. Okay, so we have an angle of pi over 6, but the sign of the angle is negative. Now in what quadrants does that take place? Well let's see. We've got sign corresponds with the notion of Y, and so we have the plus, plus, minus, minus. Ah-ha! So sign is negative in these 2 quadrants. So we have a reference angle of pi over 6 in this quadrant, and in that quadrant. Now what are the angles associated with these 2 reference angles? Alright, well we think about pi here. Now pi in terms of 6th's is 6 pi over 6, and 6 pi over 6 plus 1 more pi over 6 is 7 pi over 6. So 1 solution is 7 pi over 6. So X then is... I need a place for it here. Let's see, X is... 7 pi over 6. And then for this other one, well let's see. All the way around would be 2 pi, and then we want to subtract pi over 6. Well let's see, 2 pi in terms of 6th's would be, I'll take 2 pi and multiply by 6 over 6, that's 12 pi over 6 all the way around. Subtract 1 pi over 6, that's 11 pi over 6. So the other one is 11 pi over 6. Alright, so that's 2 of our answers. Now let's find our other answers. We march over here, sign X is plus 1 is 0. So sign X then... equals -1. Ah-ha. Now I mentioned before on a previous problem, we said that if the cosign of angle, or the sign of an angle was 0, then it's a little easier to look at the graph. And it turns out that looking at the graph is easier as well when we are either at the top or the bottom of the amplitude. We can just see it more clearly. We can see it using the circle, but it's real easy to see if we are... looking at the graph. Oops, I'm gonna do the graph differently if we're looking at the graph. Now for the sign curve, let's see. We start at an intercept, and we end up at 2 pi, and critical values are here and here, and we start here, we go to the top of the amplitude, bottom of the amplitude, back to... there we go. Alright so now, sign of what angle is -1? Well -1 occurs right here. Oh, that's at this point, that's between pi and 2 pi. That's 3 pi over 2.