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). tan 3x = (√3)/3

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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 tangent of six X equals one. Our answer choices are answer choice. A no solution answer choice B X equals the solution set. Pi divided by six pi divided by three pi divided by 22 pi divided by 34 pi divided by 33 pi divided by 25 pi divided by three and 11 pi divided by six. They get longer answer choice C X equals the solution set. Pi divided by 12 pi divided by 45 pi divided by 12 7 pi divided by 12 3 pi divided by 4, 11 pi divided by 12 13 pi divided by 12 5 pi divided by 47 pi divided by 12, 19 pi divided by 12 7 pi divided by four and pi divided by 12 answer choice D X equals the solution set. Pi divided by 24 5 pi divided by 24 3 pi divided by 8, 13 Pi divided by 24 17 pi divided by 7. Pi divided by 8 25 pi divided by 24 29 pi divided by 24 11 pi divided by 8 37 pi divided by 24 41. Pi divided by 24 15 pi divided by eight. All right. So we've got the tangent of six X equals one. Let's set aside our angle of six X for a moment. Let's set six X equal to angle A. So we're talking about the tangent of angle A equals one. Where is it true that the tangent of an angle equals one? Well, we recall from previous lessons and from the tables in the textbook that the tangent of angle A equals one when angle A equals pi divided by four. Now that's in the first quadrant. Is it true anywhere else? Well, tangent is positive in both quadrant one and in quadrant three. So in quadrant three, if we use pi divided by four as our reference angle, and we take pi plus pi divided by four, that's going to be at five pi divided by four. So we have a equal to both pi divided by four and five pi divided by four. But recall that A is not A A is six X. So we have two equations. We have that six X is equal to pi divided by four. And we have that six X is equal to five pi divided by four. But we don't want to leave it at just that we want to find where this is true all throughout our interval of 0 to 2. Pi. So what we do is we add the period. So if we take six X equals pi divided by four and we add the period, the period for a tangent is pi but we're going to add N pi because N is going to represent all of the integers, we'll use to find where it's true all throughout the interval. And then we'll also take our six X equals five pi divided by four plus our period of N P. And now we just need to solve for X. So we're going to divide everything by six for both equations. And so we'll have X equals and for our pi divided by four, that's pi divided by 24. And then to add the period of N pie, that'll be N pi divided by six. But we want to have a common denominator of 24. So that's going to be four and pi divided by 24. And very similarly dividing everything by six. For the second equation will have five pi divided by 24. And then plus, and we'll use again four and pi divided by 24 so that we have the common denominator. So there are our two equations. Now, it's just a matter of plugging different values in for N to find where this is true. So we'll start with that first equation where we've got pi divided by 24 plus four N P divided by 24. And we'll use an N value of zero. So we've got pi divided by 24 plus four, multiplied not by N but by zero, multiplied by pi all divided by 24. While that second fraction with the zero being multiplied in the numerator, that's just zero. So this simplifies to pi divided by 24. So there's our first one. Our first solution for X. Now we'll use an N value of one. And so we'll have pi divided by 24 plus four, multiplied not by N but by one multiplied by pi all divided by 24. And so four multiplied by one is four. We have four pi plus pie. That's going to be a five pie divided by 24. There's our next value for X. All right, we'll try N equals two. We have pi divided by 24 plus four, multiplied not by N but by two, multiplied by pi all divided by 24. 4, multiplied by two is eight. So we have eight pi plus pi that's nine pi divided by 24. But this one can be reduced. They're both divisible by three. So this one equals three. Pi divided by eight. There's our third solution. Now we'll try N equals three. How high are we going to go? Just pausing for a moment. If we are looking at a limit of two pie, we can't get to two pie. We want to get as close to two pie as possible without going over. But we've got denominators of 24 here. So if we take two pi multiplied by 24 divided by 24 we're looking at 48 pi divided by 24 as our limit. So we want numerators getting close to 48 as close as possible without getting to 48 or going over. So there's our goal. All right. So we got pi divided by 24 plus. Now N equals three, we've got four, multiplied by three, multiplied by pi all divided by 24 for four, multiplied by three is 12. So we have 12 pi plus one pie is 13 pi divided by 24. That's simplified. There's our next X value. Now we'll try N equals four. We've got pi divided by 24 plus four, multiplied by four instead of N multiplied by pi all divided by 24 4, multiplied by four is 16. So 16 pi plus one, pi is 17 pi all divided by 24. And that one's simplified. All right. Now, we've got an N value of five pi divided by 24 plus four, multiplied not by N but by five, multiplied by pi all divided by 24. 4, multiplied by five is 20. So 20 pi plus one, pi is 21 pies divided by 24. Those are both divisible by three. So we can simplify that to seven pi divided by eight. We're about halfway there. All right, let's try an end value of six. So we've got pi divided by 24 plus four, multiplied not by pi but by er, excuse me, not by N, but by six still multiplied by pi all divided by 4 multiplied by six is 24 plus one. Pi is 25. So 25 pi divided by 24. And there's our next X solution. OK. Now we've got an N value of seven pi divided by 24 plus four, multiplied by seven. Didn't switch colors. Four multiplied by seven, multiplied by pi all divided by 4. Multiplied by seven is 28 plus one. That's 29 pi divided by 24. That one is simplified. Now we'll try an end value of eight pi divided by 24 plus four. Multiplied not by N but by eight multiplied by pi all divided by 24 4, multiplied by eight is 32 plus one is 33 P divided by 24. Those are both divisible by three. So that simplifies 2 11 pi divided by eight. All right, got that one. Now, let's try an N of nine. We've got pi divided by 24 plus four, multiplied by nine, multiplied by pi all divided by 24 4, multiplied by nine is 36. So 36 pi pi plus pi is 37 pi divided by 24. That one's simplified and we'll try an N value of 10. We are not at 48 yet for our numerators. So we got pi divided by 24 plus four, multiplied by 10, multiplied by pi all divided by 24 4, multiplied by 10 is 40. So 40 pi plus one is 41 pi divided by 24. All right. And that one's simplifying. Now, let's try an N of 11 as we're renting out a room on this part of the screen. We've got pi divided by 24 plus four, multiplied by 11, multiplied by pi all divided by 24 4, multiplied by 11 is 44 44. Pi plus one. Pi is 45 pi all divided by 24. Those are divisible by three. So that's 15 pi divided by eight. And we've been adding, let's see how much is that for pie every single time here. If we add another one, we'll get to 49 pie divided by 24 which will put us over our limit for hour interval. And notably, we did say 1000 years ago when we started this, that we've got this second equation. Five pi divided by 24 plus four N pi divided by 24. If we put a zero N for N there, that one will be five pi divided by 24 which we had five pi divided by 24. It would just get us some of the same solutions. So we don't have to go through and plug all of those N values in. For that second equation. It'll just repeat. But there is our solution set. If we look at our answer choices, this matches with answer choice D 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). tan 3x = (√3)/3

Key Concepts

Multiple-Angle Trigonometric Equations

Properties and Periodicity of the Tangent Function

Solving Trigonometric Equations on a Restricted Interval

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