In Exercises 53–62, solve each equation on the interval [0, 2𝝅)....

Question

In Exercises 53–62, solve each equation on the interval [0, 2𝝅)._(2 cos x + √ 3) (2 sin x + 1) = 0

Accepted Answer

Hey, everyone in this problem, we're given a trigonometric equation and we're asked to solve on the interval from 0 to part including the end points. And to give the answer an exact radiant measure the equation, we have, we have two factors multiplied together, we have 10 co X minus five, multiplied by four sine X minus two, multiplied by the square root of three is equal to zero. We have four answer choices here. OK? Option A is that there is no solution. And then option BC and D are sets containing three possible X values and we'll come back to those as we work through the problem. Now, let's start just by writing the equation we have. So we have cos er sorry, 10 cosine of X minus five and all of that is multiplied by four B X minus two root three and this is equal to zero. So what we have is we have two factors multiplied together, equaling zero. OK. So if either one of these factors are equal to zero, the left hand side will be equal to zero and the equation will be satisfied. So we're gonna look at each of these individually. So we're gonna start with 10 cosine of X minus five and we're gonna set that equal to zero. And just like any other equation we have. What we wanna do is we want to try to isolate X. So we want to get all the numbers to one side and have X on the other side. We're gonna start by moving the five to the right hand side by adding cosine of X is equal to five, dividing by 10, gives us that co X is equal to one half. All right now we have coast X is equal to one half. And let's start by finding our reference angle. And so a reference angle is gonna be that angle that is less than pi over two. And we're always measured from the X axis. So we have X prime, which is gonna be our reference angle and we're gonna use one of our special triangles. OK. Special triangles I think is a really great way to be able to remember all of these angles. It makes it a lot simpler. You have less to remember just remember these triangles and you can figure out the angles that you need. So we have our triangle, we know we have some sides with one and two. So we're gonna use the following special triangle. We have a one on the shortest side, we have two on the high partners. So the other side must be the square root of three. And you can always double check your sides by Pythagorean theorem. OK. We have that one squared plus the square root of three squared is equal to four in the square root of forest two which gives us the hypotenuse. All right. Now, for our angles, the angle opposite our smallest side. OK. Our smallest side is one, the angle opposite that is gonna be the smallest angle. OK. And this triangle is gonna be pi divided by six. And then the other angle is pi divided by three. OK. So this is our special triangle. And now we can figure out what cosine of one half is. OK. For the cosine, we take the adjacent side divided by the hypotenuse. And so the angle is going to be pi divided by three here or reference angle X prime is gonna be pi divided by three. Now, we have to be careful. That's not necessarily the only answer we have. OK. We had that cosine of X is equal to one half a positive value. OK. So cosine is positive in quadrant one and quarter four. OK. So let's go ahead and draw a Cartesian plane. OK. And recall we have quadrant one on the top, right, quadrant two, quadrant three and quadrant four moving counterclockwise. And again, cosine is positive in quadrant one and quadrant four. OK? To the right of the X axis. So what we're gonna have is we're gonna have two possible answers OK. So we have the first and again, like I said, our reference angle always goes from the positive X axis or from the X axis. And so we have from the X axis up, we have pi divided by three and we also have the angle downwards. OK. So from the positive X axis downwards pi divided by three, right. So let's go ahead and label these angles as we are going to find them. So we have, I'm gonna label them in a red just so it stands up for this first angle. We're gonna call X one, the second angle we're gonna call X two. Remember that we measure our angles going counterclockwise from the positive X axis. So that angle is gonna go all the way around and that's gonna be called X two. Yeah. All right. Now X one, X one is easy. OK? X one is just gonna be pi over three. It's just that reference angle X two. Well, an entire rotation entire circle is gonna be two pi radiant but we stopped pi divided by three short of them. OK. So we have an entire rotation minus pi divided by three. OK. So X two is two pi minus pi divided by three. This is gonna give us five pie divided by three. OK? So we have two answers right now. We have pi divided by three and we have five pi divided by three. Now we're gonna move to the second fact, OK. So four sin X minus two multiplied by the square root of three. We're gonna do the exact same thing. OK? So let me write that out. We're gonna set it equal to zero or sine X minus two root three is equal to zero. We're gonna move the constant term to the right. We have that four sign X is equal to two root three, dividing by four. We get that sign of X is equal to the square root of three divided by two. And now we can use our special triangle again and we're gonna use the exact same one. And you can see that we have the squared of three and two. So we're gonna want that same triangle with those same side lengths. We have one in route three on the sides two for the hypos. And then our angle across from the square root of three side is going to be pi over three. Again, what you can see recall that sine is gonna be the opposite side divided by the hypotenuse. And so our reference angle again is going to be pi over three. OK. So we have the X prime is equal to pi divided by three. All right. So we have our reference angle. Now, we need to think about which quadrants we could have a solution. OK. In this case, sin is positive, which tells us that our answer could be in quadrant one or quadrant two. A for sign to be positive, we want to be above the X axis. So we draw this out. OK. We can have pi divided by three going from the positive X axis upwards. We can also be in quadrant too where we have pi divided by three going from the negative X axis upwards. A that reference angle is always measured from the closest X axis. All right. So our first angle here again is gonna be pi over three. OK. So we're gonna call these X three and X four. So X three is just gonna be that angle. That first angle which we know is gonna be pi divided by three. The second one is gonna go all the way around counterclockwise till it hits the angle in quadrant two. So let's write this oh X three we've said is pi divided by three and we already have that as an answer. OK. So we don't really need to worry about that. We already found pi divided by three. No X four. Recalled the straight line. It is going to be equal to pi, right. So we have pi radiance and then we're subtracting this pi over three that we measure from the negative X axis. So X four is gonna be pi minus pi divided by three. And this is gonna give us an angle of two pi divided by three. And so what we found in this problem, OK. We took each factor we set it equals to zero. We found the reference angle and then used what we knew about the sign and the cosine to figure out all the possible angles we could have. And we found that there were three correct answer choices. We have pi divided by three, two, pi divided by three and five pi divided by three. We compare this to our answer choice. We can see that D is the correct answer. Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 53–62, solve each equation on the interval [0, 2𝝅)._(2 cos x + √ 3) (2 sin x + 1) = 0

Key Concepts

Trigonometric Functions

Zero Product Property

Interval Notation

Watch next