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).__√ 3sin 2x = --------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 34 sign of two X equals 17. Our answer choices are answer choice. A the solution set for X equals pi divided by 35 pi divided by 37 pi divided by three and 11 pi divided by three answer choice B X equals the solution set pi divided by 65 pi divided by six answer choice C X equals the solution set pi divided by 5 pi divided by 12, 13 pi divided by 12 and 17 pi divided by 12. And answer choice D is no solution. All right. So the first thing we want to do here is divide both sides of the equation by 34. So that we just have the sign on the left hand side. So that will be the sign of two X equals 17 divided by 34 on the right that simplifies to one half. Now we want to pause and ask ourselves where is the sign of an angle equal to one half. And we recall from previous lessons in the tables in the book, that sign of an angle is equal to 221 half, excuse me, when that angle. So in this case, two X is equal to pi divided by six. And that's true for the first quadrant where we have a positive sign. But where else is that true? Well, sine is also positive and quadrant two. And so this angle is going to also be equal to if we use pi divided by six, as our reference angle for quadrant two will take pi minus that reference angle pi divided by six. And that's going to give us five pi divided by six. So two X is also equal to our angle is also equal to five pi divided by six. And now we've got two equations to help us solve for everywhere over that interval. Now, here's what we'll do. We set two X equals pi divided by six and two X equals five pi divided by six. And then to find where this is true for the whole interval, we are going to add the period, the period for sign is two pi, but we don't just add two pi we're going to add two N pi the N is going to allow us to check for all of the different integers. So we can find where this is true throughout the whole interval. And now, we just need to get X by itself. So for both of these equations, we're going to divide everything by two. And then on the left where we had two, X equals pi divided by six plus two N P, this is going to be X equals pi divided by 12. And then for the plus two and pi divided by two, well, those twos would cancel out, but we want to have a common denominator to add these fractions. So having that be a common denominator of 12, this would be 12 and pi divided by 12 that we're adding to our pi divided by 12. Now, for our second equation, we have X equals and five pi divided by six, divided by two is five pi divided by plus. And then again, we'll use 12 and pi divided by 12. So now we just need to put in different values for N and that will tell us where this is true throughout our intervals. So we'll start off with an N value of zero. So we have pi divided by 12 plus 12, multiplied not by N but by zero, multiplied by pi all divided by 12. Well, that second fraction where we have the zero being multiplied in the numerator, that's just zero. So where N equals zero, we get a solution of pi divided by 12 for the value of X. Now we can try N equals one. But how high are we going to go here. Well, we know that we're limited. We want to get as close to two pi as possible without hitting it or going over. So if we take two pi and find out what that is with the denominator of 12, we multiply it by 12, divided by 12. That's going to give us 24 pi divided by 12. So we want to get as close to 24 pi divided by 12 as we can without going over that. OK. So back to N equals one, we have pi divided by 12 plus 12, multiplied not by N but by one multiplied by pi all divided by 12. 12, multiplied by one is 12. So we have 12 pi plus one pie. That's 13 pi divided by 12. That's simplified and it's not over our 24 pi divided by 12. So that one works. Now let's try an end value of two. We have pi divided by 12 plus 12 multiplied not by N but by two multiplied by pi, all divided by 12. 12, multiplied by two is 24. So we have 24 pi plus one pi that's 25 pi as we said, we wanted to get as close to 24 pi as possible without hitting it or going over. So this is beyond our interval. So we can stop with that first equation. And now we can look at our second equation where X equals five pi divided by 12 plus 12, N pi divided by 12. And we'll start with an N value of zero again. So we've got five pi divided by 12 plus multiplied not by N but by zero multiplied by pi all divided by 12. We know the second fraction zero. So this is just five pi divided by 12. And now we'll try and end value of one. So we have five pi divided by 12 plus 12, multiplied not by N but by one multiplied by pi all divided by 12, 12, multiplied by one is 12. So 12 pi plus five pi is 17 pi all divided by 12. And now we've been adding 12 pie every time. If we added another 12 pi to 17 pie, that's going to put us over for an end value of two for the second equation. So we know we can stop there. And if we reorder these, this matches with the solution set for answer choice. C 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).__√ 3sin 2x = --------2

Key Concepts

Double-Angle Identities

Solving Trigonometric Equations

Unit Circle and Interval Constraints

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