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 cos² x + 3 cos x + 1 = 0

Accepted Answer

Hey, everyone. Welcome back in this problem. We have the trigonometric equation 14 co squared of A minus seven cos A minus seven 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 radiance, we have four answer choices. Options A and C both contain sets with three solutions. OK? And they're just different values for each and option B and D are sets containing four different solutions for the problem. We're gonna come back to those as we work through this problem. And let's start by writing out our equation cosine squared A minus seven cosine A minus seven is equal to zero. The first thing we're gonna notice is that this entire equation is divisible by set. OK. So let's go ahead and do that first just to simplify it, make these numbers a little smaller and easier to work with. If we divide both the left and right hand side by seven, we get two cosine squared A minus cosine A minus one is equal to zero. Now, this equation will look kind of familiar. OK. We have something squared minus something minus a constant term. This is kind of like a quadratic equation. So what we can do is substitute and we're gonna let X be equal to a cosine of a. If we do that, we can write our equation as two X squared minus X minus one is equal to zero. And now this looks exactly like a quadratic equation that we know how to solve for. And we can solve for X. Once we know the value of X, we can substitute it back into our X equals cosine a equation to figure out what the values of A are. So we have this quadratic equation, we want to use a quadratic formula to solve for X. And so we have that our A value is gonna be two, the coefficient corresponding with X squared term, our B value is going to be negative one, the coefficient corresponding with their X term. And our C value is going to be negative one that constant term at the end. OK. And you could factor if you wanted to factor this equation as well. Instead you could go ahead and factor that. Um we're gonna use a quadratic formula but either one is completely. OK. So we have X is equal to negative B plus or minus the square root of B squared minus four AC all divided by two A substituting in our A B and C values, we get one or minus. The square root of negative one squared minus four, multiplied by two, multiplied by negative one, divided by two, multiplied by two. All right, let's give ourselves some more room to simplify this. When we do, we're gonna get X is equal to one plus or minus the square root of nine divided by four. We know that the squared of minus three. And so we get X is equal to plus or minus, sorry, one plus or minus three divided by four. So we get two different X values. OK? If we add, we have one plus three divided by four, which gives us just one. If we subtract, we get X is equal to one minus three divided by four. And so that's gonna be equal to negative half. So those are our two X values and we're gonna put those in green boxes for now actually let me put one in green. So the X equals one in green and X equals negative half in Boli. And now I want to relate these back to A, OK? Remember we're trying to find values for the angle A, we want to relate these back to our cosine A equation. So from the first one, we now have the cosine of A is equal to one. Well, let's think about this. Hey, if we draw our cosine graph, you know that it starts at one goes downwards and comes back up back to one after one full cycle, which is gonna be at two py radiant. So we know that A is gonna be equal to zero to pie, et cetera. We're on the interval from 0 to 2 pi so we just have those two values, but we have to be very careful. OK. Our interval does not include two pie. It has a round bracket on two pie. So two pie is actually not going to be a solution. OK. So from this half of our equation cosine of A is equal to one, we get that A could be equal to zero. All right, moving over and let me just add one to this graph. So we're not confused. Moving over to our blue solution, we have X is equal to negative half if X is equal to negative half, that means we can write cosine of A is equal to negative one half. Now, the first thing we're gonna do is find our reference angle and we're gonna call it a prime and recall that the reference angle is just gonna be the angle between zero and pi over two that relates these two side links. OK. So let's think about a special triangle that has side links, one and two. And recall that we can write this special triangle with side link one. A hypo is two and the other side length is a square root of three. And you can always check these using pyy right there one squared plus the square root of three squared gives us four, we take the square root to get the hypotenuse and we get two. OK. So that's exactly what we want. Now, the smallest side is one across from the smallest side will be the smallest angle. And recall that this triangle is our pi divided by six pi divided by three triangle. OK. Or our 30 degree, 90 degree tri or 30 degree 60 degree triangle. So across from side length, one, we have the angle pi divided by six. And across from side length route three, we have pi divided by three. So we want cosine, we want the adjacent side divided by the hypo, the hypotenuse is two. So that's good. We want the adjacent side to be one. And so we're gonna take cosine of pi divided by three. OK. So a our reference angle, a prime is gonna be pi divided by three. That's the angle that relates one and two through the cosine. Now that we have our reference angle, we have to think about where we have solutions. OK? We have that cosine is a native and we know that cosine is negative in quadrant two in quadrant three. And remember that cosine relates X and R. So we want where that X value is negative, which is to the left of the Y axis, which is in quadrant two and three. So let's draw out what we have and find our solutions. We're gonna draw out our quadrants, we have quadrant two on the top left, quadrant three on the bottom left. And we have this reference angle of pi divided by three. So what we do is we go to quadrant two and our reference angle always goes from the X axis, the closest X axis. So it's gonna go from the negative X axis upwards pi divided by three units, that's gonna be a solution. And we also have from the negative X X was downwards to quadrant three pi divided by three radiant, which is going to be another solution. OK. So we have these two solutions. The first one, we're gonna call a one and the second one we're gonna call 82. Yeah, which is just gonna be done in. So a one, what's a one gonna be equal to? Well, we can imagine that if we take this angle all the way to the negative X axis, OK, that's gonna form a straight line. We're gonna have a complete angle of pie. Now we're not going all the way to the negative X axis. We're stopping pi divided by three radiant short. And so a one is gonna be equal to pi minus pi divided by three radiant and which gives us two pi divided by three. And that's one thing I really like to draw these diagrams out. It makes it really easy to visualize whether we should be subtracting or adding the reference angle and whether we should be doing it from pi or from two pi. All right. And a two, the second value we're gonna go all the way around. We're gonna pass the negative X axis and then we're gonna go pi divided by three radiant more. OK. So to the negative X axis, that straight line that's gonna be pi, then we're going pi divided by three radiant more. So our angle is gonna be pi plus pi divided by three, which gives us four pi divided by three. So we found three solutions. Let me write them out. We have zero, two pi divided by three and four pi divided by three. OK. Those are our three solutions for A on this interval. OK. The first one came from this cosine of A is equal to one. The second from cosine A is equal to negative a half. If we compare this to our answer choices, can I mention that there were two choices that had three solutions each? That was A and C? So we know we've narrowed it down to one of those two options. And if we look at the values in these sets, we can see that the correct answer is option. A 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 cos² x + 3 cos x + 1 = 0

Key Concepts

Quadratic Equations

Trigonometric Identities

Interval Notation

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