In Exercises 54–67, solve each equation on the interval [0, 2𝝅)....

Question

In Exercises 54–67, solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places.cos 2x = -1

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 zero, including 0 to 2 pi not including two pi employing exact values whenever feasible or providing approximations accurate to four significant figures. Our equation is the sign of five X equals negative one. Our answer choices are answer choice A X equals the solution set three pi divided by 10 7 pi divided by 10 11 pi divided by 10 3 pi divided by two and 19 pi divided by 10 answer choice B X equals the solution set three pi divided by 57 pi divided by five and nine pi divided by five. Answer choice C X equals the solution set two pi divided by 74 pi divided by 76 pi divided by pi divided by 7 10 pi divided by seven and 12 pi divided by seven and answer choice D is no solution. All right. So the first thing we want to ask ourselves is when is the sign of anything equal to negative one? And that is at the quadrant angle three pi divided by two. So we can rewrite this as five X is equal to three pi divided by two. And now to solve for all of the different times at which the sign of five X is equal to negative one, we want to also add the period here and the period of sign is two pi but we write plus two N pi because the N represents any integer and we're going to use that to help us stay within the interval, finding all of the times for which this is true for, for X within that interval. So now we just need to solve for X and then we can plug in different values for N in order to find our solution set. So we're going to divide everything by five in order to solve for X. And we get on the left hand side of the equal sign X equals on the right. We're taking three pi divided by two, divided by five, which is the same as three pi divided by two, multiplied by the reciprocal of five, which is 1/5. And that's how we'll get three pi divided by 10 plus our two. And that's not an N, it's almost an N to N pi divided by five. And that's how we get our period for two pi divided by B where B is the number in front of our X in the sine function. So we've got this set up to solve for X or excuse me to solve. We've got X solved for now, we just want to plug in different values of N and we'll find the, all of the solution set for X. We'll start off with an N value of zero. So we're going to take the three pi divided by plus the excuse me, not two plus two, multiplied not by N, but now by zero, multiplied by pi all divided by five. And for that second fraction, we've got zero being multiplied in the numerator. That's just going to be zero. So three pi divided by 10 plus zero, that's just going to be three pi divided by for our first value for X. All right. Now let's solve for a value of N equals one. So we'll have still three pi divided by 10 plus two multiplied now instead of N but by one multiplied by pi all divided by five. Now we notice here that to add these, we're going to need a common denominator. The common denominator will be 10. So we can multiply that second fraction by two divided by two. So carrying down our three pi divided by 10 plus and now doing all that multiplication two multiplied by one multiplied by pi multiplied by two is going to be four pi divided by five, multiplied by two is 10. Now we can add these and in the numerators, three pi plus four pi is seven pie and that's divided by 10. So there's our next value. Seven pi divided by 10. Now, we can test for N equals two. But how high can we go? Well, our limit is two pi we can get as close to two pi as possible without going over. And we've got a denominator of 10 here. So if we have two pi with the denominator of 10, we'll multiply it by 10, divide it by 10. That will give us 20 pi divided by 10. So there's our limit. We want to get as close to 20 pi divided by 10 as possible without going over. So we can test and equals two. We'll have three pi divided by 10 plus two, multiplied not by N but by two multiplied by pi all divided by five. And because it's divided by five, we're multiplying the second fraction by two, divided by two to get the common denominator, bring down our three pi divided by 10 plus. Now two multiplied by two, multiplied by pi multiplied by two is eight pi divided by 10 and three pi plus eight p 11 pi divided by 10. So there's our next value for X and we notice a pattern here. We're adding four pi every time and at 11 pi divided by 10, we're getting close to 20 pi divided by 10, but we're not there yet. Now, if we put three N for N, we've got three pi divided by 10 plus two multiplied by not N but by three, multiplied by pi divided by five. And multiplying that by two, divided by two. For the common denominator, we'll have three pi divided by 10 plus two. Multiplied by three, multiplied by two is pi divided by 10. Now adding three pie plus 12 pie gets us 15 pie divided by 10 and that one can be reduced. Both 15 and 10 are divisible by five. So that reduces to three pi divided by two. We're getting closer to 20 pi divided by 10. Let's plug in an end value of four. So we have three pi divided by 10 plus two, multiplied by four instead of N multiplied by pi all divided by five. So we multiply it by two divided by two for our common denominator. And that will be a three pi divided by 10 plus two, multiplied by four, multiply by two, multiplied by pi is 16 pi divided by 10. 3 pi plus 16. Pi is 19 pi divided by 10. While that is pretty close to 20 pi divided by 10 any further. And we will be going over 20 pi divided by 10. So here's our answer. We'll look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

In Exercises 54–67, solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places.cos 2x = -1

Key Concepts

Cosine Function and Its Properties

Double Angle Formulas

Solving Trigonometric Equations

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