Solve each equation in x over the interval [0, 2π) and each eq...

Question

Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.

2 cos 2x = √3

Accepted Answer

Welcome back. Everyone in this problem. We want to find the exact solutions of the equation 23 X equals the square root of two. And we are considering the interval for open bracket 02 pi closed parentheses for our answer choices. A says that it's 1/12 of pi 7/12 of pi three quarters of pi five fourths of pi 17 12th of pi and 23 12th of pi B says it's 1/6 of pie, two thirds of pie, five fourths of pie and seven fourths of pie. C says it's a third of pi five thirds of pie and seven thirds of pie. And D says it's an eighth of pi, 3/4 of pie, 7/8 of pie, four thirds of pi, five thirds of pi and 11 6th of pi. Now let's make a note here. So we're, we're talking about the formula two multiplied by the cosine of three X equals the square root of two. Well, no, this tells us that the cosine of three X must be equal to the square root of two divided by two. And at this point, we're trying to find this anger three X in other words, what possible anger the expression three X gives us the square root of two divided by two. In other words, the cosine of what possible angle gives us the square root of two divided by two. Well, we can use our unit circuit to help us since we are looking for the exact solution. Because recall that on the unit circle, on the unit circle here, OK? Where we have a circle, OK? And our circle, a unit circle that means it goes from one to negative one on both axes. OK. Then on the unit circle, mm we know that the X and Y coordinates are written in the form cost theta sine theta where cost theta is X and sine theta is Y. OK. So if we're saying that the cost of our anger equals the square root of two divided by two, it means that at some point where the X value is the square root of two divided by two, we want to figure out what that is. Well, recall that the cosine of theta or the X value is the square root of two divided by two. When our angle is 1/4 of pi, we also know that if we were to go in the opposite direction here, there's also another possible X value and that X value is when are, is equal to seven fourths of pie. So let me draw that here. So the other solution is seven fourths of pi, OK. So we know, then we know then that the core sign of 1/4 of pie is equal to the core sign of seven fourth soft pie, which is equal to the square of two divided by two. And if that's the case, then what does that tell us? Well, this angle must be equal to both of those values. And if it's equal to both of those values, right? OK. If you go to both of those, then we can say that three X, he calls 1/4 of pie and three X also equals seven fourths of pi. Now here's the thing, this is our solution on the unit circle. But generally, generally, we know then that that means three X is going to be equal to 1/4 of pi plus two pi N where N can be any integer. Because recall that the period of the sine function is two pi. So if we add two pi to our function or to our values, it will continue to repeat and give us the same answer. Likewise, that's the case for seven pi. So generally three X equals seven fourths of pi plus two PN. OK. So if that's the case, then we can solve for X in both and figure out what the actual solutions are within our interval. Now three X equals 1/4 of pi plus two pi and we can divide the right hand side by three and we'll get X to be equal to AAA 12th of pi plus two thirds of pi multiplied by N. Now here remember that our period goes from our interval is from 0 to 2 pi. So we could find possible values of N within that period because we have two thirds of pi two pi is being divided by three. So let's try to find all the values until pi on all the values that are less than two P basically. OK. So let's start with N equals zero. Now, when N equals zero, then X is going to be equal to 1/12 of pi when N equals one, OK, then X is going to be equal to 1/12 of pi plus two thirds of pi multiplied by one and the two thirds of pi multiplied by one that is going to be equal to two thirds of pi and 1/12 of pi plus two thirds of pi is going to be equal to 3/4 of pi beef, fourth soft pie. Next, when N equals two, I remember N can be any integer X is going to be equal to 1/12 of pi plus plus two thirds of pi multiplied by two. OK. So that's four thirds of pi and four thirds of pi plus 1/12 of pi equals 17 12th of pi. Now would we also have a solution for N equals three? Well, when N equals three, then X is going to be equal to 1/12 of pi plus two thirds of pi multiplied by three. Now three cancel three. So it's gonna be 1/12 of pi plus two pi which is going to be greater than two pi, that's going to be outside of the interval. So that N equals three cannot give us a solution. So it stops at N equals two. So our solutions when three X equals 1/4 of pi plus two, Pi N are 1/12 of pi or sorry. 1/12 of pi plus two thirds of pi multiplied by N. Our answers are 1/12 of pi 3/4 of pi and 17 12th of pi. Now essentially let's do the same thing for three X being equal to seven fourths of pi. Now three X equals seven fourths of pi. That means X is going to be equal to 7/12 of pi plus two, P divided by three N. So four N equals zero. Again, we know that X is going to be equal to 7, 12 soft pi for N equals one, we know that X is going to be equal to 7/12 of pi plus two thirds of pi multiplied by one which when we were to, we get five fourths of pie. OK? And when N equals two, when N equals two, then X is going to be equal to 7/12 of pi plus two thirds of pi multiplied by two. So that's 7/12 of pi plus four thirds of pi which when we work that out, we'll get 23 12th of pie. Again, the same situation happens when we let an equal three, we're gonna have 7, 12 soft pi plus two pi which is going to be outside of the interval. Therefore, there are solutions for, for X are 23 12th of pie, five fourths of pie and 7/12 of pie. These are the six solutions for X. And if we look back in our answer choices, six solutions for our equation. And if we look back, we can notice that the answer would have to be answer choice. A thanks a lot for watching everyone. I hope this video helped.

Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.

2 cos 2x = √3

Key Concepts

Double-Angle Identity for Cosine

Solving Trigonometric Equations

Interval and Angle Measurement Conventions

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