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).

cot(3θ/2) = ﹣√3

Accepted Answer

Welcome back. I am so glad you're here. And we're told for the following trigonometric equation solve over the interval from 0 to 2 pi inclusive of zero but not two pi. And give the answer in radiance. Our given equation is the cotangent of five Q divided by two equals negative one. Our answer choices are answer choice. A no solution answer choice BQ equals the solution set seven pi divided by 10 19 pi divided by 26 pi divided by 5 29 pi divided by 2017 pi divided by 10 and 39 pi divided by answer trace CQ equals the solution set three pi divided by 10, 14 pi divided by 15, 29 pi divided by 30 49 pi divided by 30. And answer choice DQ 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. All right. So we've got this equation. We've got cotangent of five Q divided by two equals negative one. Now let's just substitute for a moment. We'll say that angle A can be five Q divided by two. So we're just thinking about the cotangent of angle A for a moment when is the cotangent of an angle equal to negative one. Well, we recall from previous lessons that the cotangent of an angle is equal to one divided by the tangent of that angle. And so we can substitute that in here, we're saying one divided by the tangent of angle A is equal to negative one. And now we can simplify for the tangent so that it's not in the denominator, we'll multiply both sides of the equation by the tangent of angle A. So we're talking about the negative tangent of A being equal to one. Well, we know that the positive tangent of A is equal to one when A equals pi divided by four, that's in the first quadrant and tangent is negative in the 2nd and 4th quadrant. So let's just look at the second quadrant because that one comes first. So the corresponding angle in quadrant two, if pi divided by four is a reference angle, then in quadrant two, we're going to be discussing the rep pi divided by four. So we have the negative tangent of angle A equals one at three pi divided by four. So that's our A. So we can go back and we can say that angle A which we said is five, Q divided by two is equal to the pie divided by four. So now we know that we can solve for QE. But what we also need to do is we need to add our period and find out where this is true for Q across the board. So we set five Q divided by two, equal to three pi divided by four. And then we add the period, the period for cotangent is pi and we're going to take N multiplied by pi pi or excuse me, N is that integer that we're going to use to find all of the possible solutions within our interval from 0 to 2 pi. So now let's get Q by itself, we are going to multiply everything by two. Let's just do this one step at a time. So on the left-hand side, we'll have five Q that's equal to and we'll have six pi divided by four plus two NP. And now we're going to divide everything by five to get Q by itself. So we'll have Q equals and then it will be six pi divided by 20 plus two N pi divided by five. Now we can simplify six pi divided by 20. And then we'll want to change two N pi divided by five. So it has a common denominator. So six pi divided by 20 will simplify 23 pi divided by 10. And then we want a common denominator of 10. So we're going to multiply two N pi divided by five by two, divided by two. So that will be plus four and pi divided by 10. So that's what we're dealing with. Now, we just need to plug in for N, we can start off by plugging in N equals zero. So we'll have three pi divided by 10 plus four, multiplied not by N but by zero, multiplied by pi divided by 10. And that second fraction with the zero being multiplied in the numerator. That's just zero. So this is the three pi divided by 10. That's our first solution for Q. All right. Now let's plug in for N equals one. So we'll have three pi divided by 10 plus four, multiplied not by N, but now multiplied by one, multiplied by pi all divided by 10. And simplifying in the numerator, four multiplied by one multiplied by pi is four pi. So we're taking three pi plus four pi which is seven pi still divided by 10. There is our second solution for Q and now let's keep going. We'll try for N value of two. We have three pi divided by 10 plus four, multiplied not by N but by two multiplied by pi all divided by 10 and four multiplied by two. And the numerator of the second fraction is eight. So three pi plus eight pi is going to be pi divided by 10. There's our next solution for Q. Now we can try an end value of three. Now, how high are we gonna go? We have this interval, we can only get up to two pi we've got a denominator of 10. So two pi with the denominator of 10, multiplying that by 10, divided by 10 will give us 20 pi divided by 10. So that's our maximum. We can't get to that. We can get as close to that as we possibly can without hitting it or going over. So 11 pi divided by 10, we're almost halfway there. Now, for N value of three, we've got three pi divided by 10 plus four, multiplied not by N but by three multiplied by pi all still divided by 10. In the numerator form multiplied by three is 12. So we have 12 pi plus three pi is 15 pi divided by 10. But that one can be reduced. Both 15 and 10 are divisible by five. So that will be three pi divided by two. And now we can put in an end value of four. We've got three pi divided by 10 plus four, multiplied not by N but by four multiplied by pi, all still divided by 10. 4, multiplied by four is 16. So we have 16 pi plus three pi that's 19 pi divided by 10. And that's pretty close to 20 pi divided by 10. So that's as close as we're going to get. So there's our solution set for Q, we look at our answer choices and 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).
cot(3θ/2) = ﹣√3

Key Concepts

Multiple-Angle Trigonometric Equations

Cotangent Function and Its Properties

Solving Trigonometric Equations on a Restricted Interval

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