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. 3 tan² x - tan x - 2 = 0

Accepted Answer

Welcome back. I am so glad you're here. We're 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 of two pi employing exact values whenever feasible or providing approximations accurate to four significant figures. Our equation is seven tangent squared X plus six tangent X minus 13 equals zero. Our answer choices are answer choice A X equals the solution set pi divided by 67 pi divided by 64.218 and 7.360. Answer choice B X equals the solution set pi divided by 45 pi divided by 42. and 5.206. Answer choice C X equals the solution set pi divided by 34 pi divided by three and 6.283. And answer choice D is no solution. All right, looking at our equation here, we've got a tangent squared X, we've got a tangent X and a constant term. So this is a lovely quadratic. What we can do is set this up so that it's a little more intuitive to solve. We can use substitution and set the tangent of X equal to any variable. I'm going to choose U and then substituting in for the tangent of X. Now we've got seven multiplied not by tangent squared of X, but by U squared plus six multiplied not by the tangent of X, but by U minus 13 equals zero. And now we can factor this. I'm going to factor it very quickly. But if you need a refresher on factoring, definitely check out previous lesson videos, set up two open sets of parentheses. And this will factor to B seven U plus in one set of parentheses multiplied by the quantity of U minus one. And now we can solve for you two factors multiplied together equal to zero. Setting both factors equal 20. So seven U plus 13 equals zero and U minus one equals zero. Solving for you, we can subtract 13 from both sides of the first equation and we have seven U equals and negative 13 then divide both sides by seven and we get U equals negative 7th and solving for you. In the other equation, we add one to both sides and we get U equals one. So we have these two U values but we're not trying to solve for you. We are trying to solve for X. We know that U is equal to the tangent of X. So we rewrite these as the tangent of X equals negative 13 7th and the tangent of X equals positive one. So now it's just a matter of solving for X, we'll look at the tangent of X equals negative 13 7th. 1st. This one isn't something that we're going to get an exact value for. This is something we'll need our calculator to solve for X. And we do that by putting in for X equal two, we'll put in our calculator, the inverse tangent of our negative 13 7th. Make sure that you're in radiance when you plug that in and what you'll get is negative 1.768. However, that is outside of our interval, we're supposed to be between zero and two pi. So to find a co terminal angle, we'll just add two pie to that. And that will give us 5. and where is 5.2063. Well, if we recall for a moment, a coordinate plane, we have a vertical Y axis, a horizontal X axis. They come together at the origin in the middle. We know that in radiance, our left hand side of the X axis, the negative part of the X axis, that angle, that straight line is pi which is about 3.14 and so straight up on the Y axis, the positive part of the Y axis is about half of that about 1.5 and then straight down on the Y axis, the negative part of the Y axis is about 4.7. So if we've got 5.2, that's going to be in the fourth quadrant. And so we're talking about 1/4 quadrant angle here. And that makes sense because we are taking the tangent of X equal to a negative number. And so where else is tangent negative? Well, that's going to be in the second quadrant. So to figure out where our second quadrant angle is what we can do is figure out our reference angle, the reference angle is going to be the absolute value of that original angle that we came up with. So it's going to be the absolute value of the negative 1.68 or if you're putting it into your calculator, the absolute value of the inverse tangent of negative 7th. And so once we have a reference angle to get our second quadrant angle, we're going to take pi minus that. So pi minus the absolute value of our negative 1.768 gives us 2.647. So there are our 4th and 2nd quadrant angles. Now it says we need to provide these approximate approximations accurate to four significant figures. So for 5.2063, the fourth significant figure is the six that looks to the three, it stays a six. So we have 5. for that angle. And then for 2.647, the four is the fourth significant figure. It looks to the seven which tells it to round up. So for that one, we have 2.65, there are two of our angles. And now we need to take a look at that second equation where the tangent of X equals one. Now, for the tangent of X equals one, we ask ourselves, where does the tangent of an angle equal one? And that is going to be at where X equals Pi divided by four and pi divided by four. That's a first quadrant angle. And so where else is tangent positive? Well, also in the third quadrant and so what is our corresponding angle to pi divided by four in the third quadrant? Well, to get that, we're going to add pi, so we'll take pi plus pi divided by four, which will give us five pi divided by four. And there is our other angle and our other solution for X. So we look at our answer choices and this 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. 3 tan² x - tan x - 2 = 0

Key Concepts

Solving Quadratic Equations in Trigonometric Functions

Properties and Periodicity of the Tangent Function

Finding Exact and Approximate Solutions

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