Exercises 39–52 involve trigonometric equations quadratic in form...

Question

Exercises 39–52 involve trigonometric equations quadratic in form. Solve each equation on the interval [0, 2𝝅).

2 sin² x - sin x - 1 = 0

Accepted Answer

Hey, everyone in this problem, we have the trigonometric equation six sine squared P plus three sine P minus three is equal to zero. And we're asked to use algebraic techniques to solve in the interval from 0 to 2 pi excluding two pi. And to give the answer in for radiance, we have four answer choices. Options A and C are both sets containing two options for P A and option B and D are sets containing three values for P. We're gonna come back to those in more detail once we work through this problem. So let's start by writing the equation we were given, we have six sine squared P plus three, nine P minus three equals zero. First thing we're gonna notice is that the entire equation is divisible by three. So we're gonna divide the left hand side and the right hand side by three in order to simplify, just have smaller numbers to deal with. When we do that, we get two squared P plus sine P minus one is equal to zero. And this looks kind of familiar can we have a sign squared? We have a sign term, we have a constant term, this is kind of like a quadratic equation. What we're gonna do is we're gonna use substitution. OK. We're gonna let X be equal to sin A B so that we can write this as a quadratic equation in the form we're used to seeing OK, solve for X and then we can come back and solve for P. So writing our equation in terms of X is equal to sine P, gonna be two X squared plus X minus one is equal to zero. We want to solve for X. We can do that in two ways. We can either use the quadratic formula or we can factor for this one. Let's go ahead and factor. And when you factor you can factor in whatever way you like to do most, I'm gonna choose to split our middle term or our linear term A in order to factor. So we're gonna write our equation as two X squared plus two X minus X minus one is equal to zero. And then we can factor two terms at a time. So from the first two terms, we have two X squared plus two X, we can factor two X, we're left with X plus one in the second two terms. And we have this negative one that we can factor as a common factor. We're left with Xbox one, all of this is equal to zero. And now we have this X plus one term. So we can factor that out, we're left with two X minus one. So we have X plus one multiplied by two X minus one is equal to zero. That is our factored form. And this gives us two possible X values. So the first X value we get, it's gonna be X is equal to negative one. And the second X value we get is that X is equal to one. If either of these factors are zero, the left hand side is equal to zero. And that's where those X values come from by setting those each equal to zero. So we have these two possible X values and we're gonna work through a solution for P for each of them. OK. Remember we're looking for P, we know that X is equal to sine P, we have these X values. So from the first one, this tells us that sign of P is equal to negative one. OK? Well, sign A P is equal to negative one between zero and two. Pi let's recall a sketch of our sine graph. It starts at 00, it goes upwards back downwards to its minimum value and then back to zero and it crosses the axis at pi and at two pi OK. Also at zero. But that's where we start. OK? It has this maximum at one and this minimum at negative one. So this minimum point is the point we're interested in. And we can see from our graph that occurs between pi and two pi. And so that's gonna occur at three pi divided by two. Hm. So we get that P is equal to three pi divided by two as one of our solutions. OK. This is a really handy way to work through some of these trig equations. OK. If you can remember what the graph of sine and cosine look like, it can be a great way to pick off particular values. All right. Now, the second half of this, we found that sign of P could also be equal to one half. All right. So let's think about our reference angle first. OK. We have sin of P is equal to one half. Our reference angle is gonna be an angle between zero and pi over two. But we measure from the X axis. In this case, we have sin of P is positive. So this is going to be a solution but maybe not the only solution. So we wanna think about our special triangle that has sides of one and two. And recall that we have the triangle with sides one square root of three and with hypotenuse two. And you can check those side links if you forget using Pythagorean now is gonna be the opposite side divided by the hypo. So in order to have our opposite side divided by our hop hypotenuse to be one divided by two, we need to take the angle opposite of the side length of one since one is the smaller angle in our special triangle, that angle is gonna be pi divided by six, the smaller angle. OK. And this again, this is a special triangle. There are two of them. And if you remember the values in these triangles, it can really, really help dealing with these values. All right. So we have our reference angle. Now we're gonna call it P prime. And using our special triangle, we found that it was pi divided by six. No, that's not our only answer. OK. We know that sign is positive in quadrant one and quadrant two. OK. Sine relates the Y value and the R value. So as long as Y is positive, as long as we're above the X axis sine is positive. So we get that quadrant one and quadrant two. So what we're gonna do is draw out our quadrants. OK? And we don't really need to worry about the bottom ones. We know that sin is negative there. In quadrant one, we have our reference angle measured from the X axis of pi divided by six. And that's gonna be a solution. OK. P one is gonna be pi divided by six because we measure from that positive X axis in Q two, our reference angle is measured from the nearest X axis. So we're measuring from the negative X axis pi divided by six radiant. If we want to write this as a solution, we want to write the angle from the positive x axis all the way around to this angle that we're interested in. OK. And we're gonna call this P two. And you can see that if we carry P two, this line all the way down to the negative X axis, that's gonna be pi radiance. But we stop pi divided by six radiant short. So P two is actually gonna be pi minus pi divided by six, which we can write as five pi divided by six. And those are the two solutions in that case. And so what we found is that we have three solutions in total and we have the one solution we found from sine P being negative one. And we have the two solutions we found from sine P being equal to one half. If we put these together, we can write our set as pi divided by six, five pi divided by six and three pi divided by two. And that is gonna be the full solution set for this equation on the given interval. If we compare this to our answer choices, OK? We can already eliminate options A and C because they only have two values, we know that we have three values in our set. Option B has the correct number of values, but the values are different than what we found. Those are not the correct values. And so what we see is that we have D as the correct answer. Thanks everyone for watching, I hope this video helped see you in the next one.

Exercises 39–52 involve trigonometric equations quadratic in form. Solve each equation on the interval [0, 2𝝅). 2 sin² x - sin x - 1 = 0

Key Concepts

Quadratic Equations in Trigonometry

Trigonometric Identities

Interval Notation and Solutions

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