Exercises 25–38 involve equations with multiple angles. Solve ...

Question

Exercises 25–38 involve equations with multiple angles. Solve each equation on the interval [0, 2𝝅). sec(3θ/2) = - 2

Accepted Answer

Welcome back. I am so glad you're here. We are told for the following trigonometric equation solve over the interval from 0 to 2 pi inclusive of zero but not of two pi and give the answer in radiance. Our equation is the C can of five A divided by two equals negative two square at three divided by three. Our answer choices are answer choice. A A equals the solution set. Pi divided by 37 pi divided by 15 17 pi divided by 15, 19 pi divided by 15 and 29 pi divided by answer choice B A equals the solution set pi divided by three and seven pi divided by 15 answer choice C A equals the solution set pi divided by 12 pi divided by eight pi divided by six pi divided by three pi divided by 23 pi divided by four pi two pi divided by 35 pi divided by 67 pi divided by 63 pi divided by two and 11 pi divided by six. In answer choice D a little bit shorter is no solution. All right. So taking the C can of five A divided by two let's set five a divided by two to the side for a moment. Let's set that equal to X. So we're talking about the C can of X instead of five A divided by two, still equal to negative two square at three divided by three. And we recall from previous lessons that the C can of X is equal to one divided by the cosine of X. So we can just plug one divided by the cosine of X in for the CC of X setting that equal to negative two square at three divided by three. Now we have two fractions set equal to each other. And we know that it's true that the reciprocals of those fractions are equal to each other. So we can say that the cosine of X equals the reciprocal of the right hand side is negative three divided by two square at three. Now we want to rationalize this get that radical out of the denominator. So we multiply both the numerator and the denominator by the square root of three. And in the numerator, we have negative three square at three that's divided by, in the denominator, two multiplied by the squared of three, multiplied by the squared of three is two multiplied by three, which is six. And that negative three in the numerator and the six in the denominator reduce each other. So this will just be negative square at three divided by two. All right. So we are talking about the cosine of X equal to negative square at three divided by two. And we can think for a moment about the cosine of X when is that equal to the positive square at three divided by two? And we know that that is true when X is equal to pi divided by six, which is in quadrant one. And so where would cosine be negative? Well, we know that cosine is negative in quadrants two and three. So if we use pi divided by six as our reference angle, we can solve for the cosine of X for quadrants two and three. So for quadrant two, we take pi minus our reference angle pi divided by six. And that's going to give us an X value equal to five pi divided by six. There's our quadrant two. Now for quadrant three, we're going to take pi plus our reference angle pi divided by six and that will give us an X value of seven pi divided by six. Now remembering that X is equal to five A divided by two. So we have two equations here, we have that five A divided by two equals five pi divided by six. And we have that five A divided by two equals seven pi divided by six. Now we want to solve for a for our angle, but we want to solve for a anywhere within our interval, 0 to 2 pi. So to do that, we're going to add the period, the period for cosine and C can is two pi so we're going to add not just two pi but two N pi N is going to represent all of the integers that we'll use to find the values for a, within that interval. And so we add two N pi to both of those. Now we just need to solve for A and then we'll find our solution set for each, including both of these two equations. So let's start off solving for A. We're going to multiply everything by two for our equation five A divided by two equals five pi divided by six plus two. And pi. So that will leave us with five A equals 10 pi divided by six plus four. And Pi and the next step to solve for A is to divide everything by five. And then I'm not going to write a equals, but that will give us for the right hand side. It will give us 10 pi divided by plus four and pi divided by five. Now, we can reduce that first fraction 10 pi divided by 30. That would just be pi divided by three plus four N pi divided by five. And that's great. But we want a common denominator before we start plugging in for our different integers for N. So let's get our common denominator of 15. So this will be five pi divided by 15 plus and pi divided by 15. And that's the one we'll use plugging in our different integers. So putting a box around that one and for the first integer will test will be an N of zero. So for this one, just for this one, I'm going to use this equation where we simplified it pi divided by three plus four, N pi divided by five. So we've got pi divided by three plus four, multiplied not by N but by zero, multiplied by pi all divided by five. The second fraction with the zero being multiplied in the numerator. That's just zero. So that one simplifies to just pi divided by three. So there's our first solution for a. Now we'll try with an N value of one using our five pi divided by so that we can add these fractions plus multiplied not by N but by one multiplied by pi divided by 15 in the numerator of the second fraction, 12 multiplied by one is 12. So that's 12 pi plus five pi that's going to be pi divided by 15. And that is simplified 17 pi divided by cannot be further reduced. Now, that is another of our solutions. But how high are we going to go here? Well, we said before that we've got a limit of two pie. We can get as close to two pie as possible without going over. We've got a denominator of 15. So what would two pie be with a denominator of 15. So multiplying it by 15, divided by 15, that's going to be a 30 pie divided by 15. So we want to watch for a numerator getting close to 30 when we have our denominator of 15. Let's try the next end value of two. We've got five pi divided by 15 plus 12, multiplied not by N but by two multiplied by pi all divided by 15, 12, multiplied by two is 24. So 24 pi plus five pi well, that is 29 pi divided by 15. That's as close to 30 pi divided by 15 as we're gonna get before hitting it or going over. So there is the third solution and we can stop after N equals two for the first equation. But we've got another equation. Sometimes the second equation gets us repeat answers. But in this time because one was in quadrant two and one's in quadrant three, we're going to get unique answers. So we've got to keep going. We now need to solve for this equation for A. So we've got five A divided by two equals seven pi divided by six plus two N P. We're going to multiply everything by two and then we have five A equals 14 pi divided by six plus four N P. Now we're going to divide everything by five and we get R A and we're not going to write the A equals and we'll have 14 pi divided by 30 plus four N pi divided by five, we can reduce 14 pi divided by 30. Those are both divisible by two. So that will be a seven pi divided by plus. And then we can take our four N pi divided by five and then get that to a common denominator 15, which will be again, 12 N pi divided by 15. So using this one, we'll start with an end value of zero again. And we know when we take seven pi divided by plus 12 multiplied not by N but by zero, multiplied by pi all divided by 15. The second fraction is just zero. So this one is seven P divided by 15. So there's one more solution for our A. Now we can try and end value of one over here. Seven pi divided by 15 plus 12 multiplied not by N but by one multiplied by pi all divided by 15. 12, multiplied by one is 12 plus 12. Pi plus seven P is going to be 19 pi divided by 15. There's another solution for our A and now we'll try and end value of two. We have seven pi divided by 15 plus 12, multiplied not by N but by two multiplied by pi all divided by 15 and 12, multiplied by two is 24. So 24 pi plus seven pi is going to get us to 31 pi divided by 15. And as we said, before our limit was getting as close to 30 pi divided by 15 as possible without hitting it or going over. So this one goes over, that one does not get included for our solution set. Now, if we rearrange these in order our solution set matches with answer choice. A well done, we'll catch you on the next one.

Exercises 25–38 involve equations with multiple angles. Solve each equation on the interval [0, 2𝝅). sec(3θ/2) = - 2

Key Concepts

Multiple-Angle Trigonometric Equations

Secant Function and Its Properties

Solving Trigonometric Equations on a Restricted Interval

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