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

Question

Exercises 25–38 involve equations with multiple angles. Solve each equation on the interval [0, 2𝝅).                       __                     √ 3cos 4x = ﹣ ---------                       2

Accepted Answer

Welcome back. I am so glad you're here. We're 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 trigonometric equation is the cosine of three G equals negative one half. Our answer choices are answer choice A G equals the solution set pi divided by six and pi divided by three answer choice B G equals the solution set two pi divided by 94 pi divided by 98 pi divided by 9 pi divided by 9 14 pi divided by nine and 16 pi divided by nine. Answer choice C G equals the solution set pi divided by 12 pi divided by eight pi divided by four pi divided by two and pi and answer choice D is no solution. All right. So we've got this cosine of an angle which is three G is equal to negative one half. If we think for a moment when cosine of any angle, we can say beta is equal to, when's it equal to a positive one half? Well, that's when we recall from previous lessons. And from the tables in the book, that's when theta is equal to pi divided by three. So that's the quadrant one angle. But here we're talking about an angle that's equal to a negative one half. Well, where is cosine negative? Well, cosine is negative in quadrants two and three. So if we're talking about it being equal to negative value for quadrant two, we're going to take pi minus the reference angle from quadrant one which is pi divided by three and pi minus pi divided by three is two pi divided by three. That's our quadrant two angle. And how about four quadrant three? Well, to find that we take pi plus our reference angle which is pi divided by three and that's going to give us four pi divided by three. So those are the two angles here, but we don't have theta, we have three G. So we're told that three G equals both two pi divided by three and three G equals four pi divided by three. Now, we want to figure out where this is true for our entire interval from 0 to 2 pi. Now, in order to do that, we need to get G by itself and we need to add the period, the period for cosine is two pi, but we're not just going to add two pi, we're going to add two N pi that N is representing all of the different integers that we'll use in order to get all of the solutions over our interval. So we have three G equals two pi divided by three plus two N pi and three G equals four pi divided by three plus two N pi. Now let's get G by itself. For both of these, we're going to divide everything by three. And so we'll have G equals for two pi divided by three, divided by three. That's going to be a two pi divided by nine plus for two N pi divided by three. That's two N pi divided by three. But we want a common denominator of nine. So multiplying the numerator and the denominator by three will have six N pi divided by nine. And now same thing with the second equation dividing everything by three, we have G equals and we'll have four pi divided by nine plus. And we'll use that same so that we have the common denominator six N pi divided by nine. So now we're going to use those two formulas and we're going to use different integers to find where this is true throughout that whole interval. So first, we'll use an N value of zero plugging in zero for two pi divided by nine plus six, multiplied not by N but now by zero, multiplied by pi all divided by nine. All right, when we are multiplying by zero in the numerator, that's just zero for that second fraction. So this is going to be two pi divided by nine. So there's our first one. Now, let's try and end value of one. So we have two pi divided by nine plus six, multiplied not by N but by one multiplied by pi all divided by 96, multiplied by one is six. So we have two pi plus six pi which is eight pi. So we have eight pi divided by nine. And there's our second solution. Now, how high are we going to go? We have the interval going all the way up to two pi but we have a denominator of nine. So if we take two pi and multiply that by nine divided by nine, that's going to be 18 pi divided by nine. But we can't, that's not inclusive. So we want to get to as close to 18 pi divided by nine as possible without hitting it or going over. We're not there yet with eight pi divided by nine. So let's keep going. We'll try an end value of two. So we've got two pi divided by nine plus six, multiplied not by N but by two, multiplied by pi all divided by 96, multiplied by two is 12, 12 pi plus two. Pi is 14 pi all divided by nine. Now we've been adding six pie to the numerator every time. If we add six pie to the 14 pie, that's going to put us at 20 pie. So for an end value of three, that would put us over our limit. Of 18 pie divided by nine. So we can stop there for that one. Now we'll do the second equation. We'll use an end value of zero. So we've got four, excuse me, not a two. Now we're on to four pi so four pi divided by nine plus six, multiplied not by N but by zero, multiplied by pi all divided by nine. We know the second fraction is zero. So this is just four pi divided by nine. Now we'll use an N value of one. So we have four pi divided by nine plus six, multiplied not by N but by one multiplied by pi all divided by 96, multiplied by one is still six. So four pi plus six pi is 10 pi. So 10 pi divided by nine for that one and an end value of two, we have four pi divided by nine plus six, multiplied not by N but by two multiplied by pi all divided by 96, multiplied by two is 12. So 12 pi plus four pi is 16 pi. So 16 pi divided by nine and adding another six pi to that numerator would put us over our limit of 18 pi divided by nine. So this is our solution here. Two pi divided by nine, then four pi divided by nine, then eight pi divided by nine, then 10 pi divided by nine, then 14 pi divided by nine and 16 pi divided by nine. We look at our answer choices and that matches with answer choice. B 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). __ √ 3cos 4x = ﹣ --------- 2

Key Concepts

Multiple Angle Formulas

Solving Trigonometric Equations

Interval Notation

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