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).x ___tan ------- = √ 32

Accepted Answer

Welcome back. I am so glad you're here. We're 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 X divided by two equals one. Our answer choices are answer choice A X equals pi divided by eight answer choice B X equals pi divided by four answer choice C X equals pi divided by two and answer choice D no solution. All right. So we've got the tangent of X divided by two for just a moment. Let's set aside this X divided by two angle. Let's say that that's a. So we're talking about the tangent of angle A being equal to one. Where is it true that the tangent of an angle is equal to one? Well, we recall from previous lessons that that is true when the angle is equal to pi divided by four, which is in quadrant one. Now, where else is tangent positive? It's also positive and quadrant three. And so we can take our angle from quadrant one pi divided by four. That can be our reference angle. And we can add pi plus our reference angle pi divided by four. And that will give us five pi divided by four. That's our angle in quadrant three. So we have angle A equal to pi divided by four and five pi divided by four. But we're calling that angle A is actually X divided by two. So we can set X divided by two equal to pi divided by four and we can set X divided by two, equal to five pi divided by four. But we want to find out where X is true for the entire interval from 0 to 2 pi. So what we want to do is add the period, the period for tangent is pi. But we're going to add, not just pi, we're going to add N pi because N is going to represent the integers. We'll use to find out where this is true throughout that whole interval. So we've got two equations. X divided by two equals pi divided by four plus N pi and X divided by two equals five pi divided by four plus. And pi we want to solve four X for both. So we're going to multiply everything by two. And so for the first equation, we'll have X equals and two pi divided by four would simplify two pi divided by two plus and then we'll have two N pi but to get a common denominator of two, that's going to be four and pi divided by two. Now we'll work on the second equation at the same time just multiplying everything by two. And so we'd have X equals and that's going to be a five pi divided by two plus. And then we'll use again four and pi divided by two. So that we have a common denominator to add those fractions. So now we'll test different values for N, we'll start with an N value of zero for the two equations. And we want to make sure that we're within this interval if we want to get as close to two pi as possible without going over it or hitting it. What is our limit if we've got a denominator of two? So we, we take two pi multiplied by two divided by two. We're looking at four pi divided by two as our limit. So now let's test for N equals zero in our first equation. So we'll have pi divided by two plus four multiplied not by N but by zero multiplied by pi all divided by two. Well, that second fraction with zero being multiplied in the numerator. That's just zero. So this one is pi divided by two. That's our first solution for X. Now, we can plug in a one for N, we test for N equals one. And that will give us pi divided by two plus four, multiplied not by N, but by one multiplied by pi all divided by 24, multiplied by one is four. So we have four pi plus one pi that's five pi divided by two. But that's over our four pi divided by two. That's more than two pi it's outside of our intervals. So that one gets disregarded. Now, looking at our second equation, well, we know when we plug in an end value of zero, 45 pi divided by two plus four, multiplied no longer by N but by zero, multiplied by pi all divided by two. That second fraction is just going to be zero. And we'll just have five pi divided by two. But we just said five pi divided by two exceeds our limit of four pi divided by two. So the only solution here is that X equals pi divided by two which matches with 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).x ___tan ------- = √ 32

Key Concepts

Solving Trigonometric Equations

Tangent Function and Its Values

Multiple-Angle Equations and Interval Restrictions

Watch next