In Exercises 97–116, use the most appropriate method to solve ...

Question

In Exercises 97–116, use the most appropriate method to solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places. 2 cos 2x + 1 = 0

Accepted Answer

Welcome back. I am so glad you're here. We are asked to apply the most suitable technique to determine the solution of the following equation within the interval from 0 to 2 pi inclusive of zero but not two pi employing exact values whenever feasible or providing approximations accurate to four significant figures. Our equation is two cosine two X minus square at three equals zero. Our answer choices are answer choice A X equals the solution set. Pi divided by 6 11 pi divided by 6 13 pi divided by six and 23 pi divided by six answer choice B X equals the solution set pi divided by 11 pi divided by 12 13 pi divided by 12 and pi divided by 12 answer choice C X equals the solution set five pi divided by 67 pi divided by 13 pi divided by 12 and 19 pi divided by and answer choice D X equals the solution set 05 pi divided by 12 7 pi divided by and two pi. All right. So first, we want to get our cosine of two X by itself on one side of the equation. So we can start off by adding the square of three to both sides of the equation. And then we'd have two cosine two X equals the square root of three. And now we can divide both sides by two to get rid of that two in front of our cosine two X. So we'll have on the left, the cosine of two X equals on the right, the squared of three divided by two. And now for the moment, we can set this two X to the side, let's say that two X equals our angle A. So we're talking about the cosine of A equals the square it of three divided by two. Now when is it true that the cosine of an angle is equal to the square root of three divided by two, we recall from previous lessons and from tables in the book that this is true when A equals pi divided by six. Now that is in the first quadrant. Is it true anywhere else? Well, yes, cosine is also positive in quadrant four. And if we use pi divided by six as our reference angle, what would be the corresponding angle here in quadrant four? Well, if we take two pi minus pi divided by six, that's going to give us 11 pi divided by six. So that is our other angle. Now instead of co angle A, remember that A is two X. So we can say that two X equals pi divided by six. And that two acts equals 11 pi divided by six. And we want to solve for X but we want it to solve for X everywhere within our interval from 0 to 2 pi not including two pi. And so what we do is we add the period, the period for cosine is two pi and we don't add just two pi, we add two N pi and that N represents all of the integers that we're going to use to find where this is true within our interval. So we do still need to get X by itself. So we're going to be dividing both sides by two or both equations. And then, so we'll have an X equals, I'm not going to write the X equals part, but we'll have pi divided by 12 plus two N pi divided by two for this first one. And we want to have a common denominator when we're adding these. So getting the second fractions denominator to be 12, we'll have pi divided by 12 plus 12 N pi divided by 12. So we'll use that in a moment. Let's simplify our second equation where we were going to divide everything by two. And so for that one, we'll have 11 pie divided by 12 plus. And again, that would be the two N pi divided by two. And we know that that will need to have the common denominator of 12. So that will be 12 and pi divided by 12. So that's our second equation. And now let's start off with an end value of zero. For our first equation. Pi divided by 12 plus 12. N pi divided by 12. So we'll have pi divided by 12 plus 12, multiplied not by N but by zero, multiplied by pi divided by 12. While that second fraction with the zero being multiplied in the numerator. That's just nothing, that's zero. So this one will be pi divided by 12. And that is the first solution for X pi divided by 12. Now we'll try an end value of one. So we'll 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 and the numerator is 12. So we have 12 pi plus pi that's going to be 13 pi divided by 12. And there is a second solution for X. But how high are we going to go here? Well, we mentioned before that we want to get as close to two pi as possible without hitting it or going over if we've got a denominator of 12, what's that going to be? So we have two pi multiplied by 12, divided by 12. That's a 24 pi divided by 12. So we want to get as close to 24 pi divided by 12 as possible without going over. So let's check. Now, a value of N equals two. We'll have pi divided by 12 plus 12, multiplied not by N but by two multiply by pi all divided by 12, 12, multiply by two is 24. So 24 pi plus pi that's pi divided by 12. That goes over our 24 pi divided by 12. That's not within our interval. So we don't need that one. Now, we can switch to checking our second equation. So we'll use an N value of zero in our 11 pi divided by 12 plus 12, multiplied not by N but by zero multiplied by pi divided by 12. We know that that second fraction here is just zero. So this one is 11 pi divided by and that one's within our interval, that one's fine. So now let's try and end value of one for the second equation. That'll be 11 pi divided by 12 plus multiplied not by N but by one multiplied by pi all divided by and 12 multiply by one is 12. So that's 12 pi plus 11 pie. That's going to be 23 pie all divided by 12. And that is as close as we're going to get to that 24 pi divided by 12 without hitting it or going over. So no need to test for N equals two for the second equation. Now, if we put those in order, we will get a solution set that matches with answer choice B well done we'll catch you on the next one.

In Exercises 97–116, use the most appropriate method to solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places. 2 cos 2x + 1 = 0

Key Concepts

Double-Angle Identity for Cosine

Solving Trigonometric Equations

Interval Restriction and Exact vs Approximate Solutions

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