In Exercises 121–126, solve each equation on the interval [0, ...

Question

In Exercises 121–126, solve each equation on the interval [0, 2𝝅). 10 cos² x + 3 sin x - 9 = 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 14 cosine squared X plus three sine X minus 12 equals zero. Our answer choices are answer choice. A the solution set X equals pi divided by 65 pi divided by 63.431 and 5.993. Answer choice BX equals the solution set pi divided by 32 pi divided by 32.852 and 5.993. Answer choice CX equals the solution set pi divided by 65 pi divided by six and 0.2898. And answer choice D is no solution. All right. So what have we got here? We've got a cosine squared X. We've got a sine X and then we have a constant term and we want to get these all to be like terms. So we can work with this So we recall from previous lessons, the identity that cosine squared X. What we have here is equal to one minus sine squared X. So now we can use substitution for the cosine squared X. We can substitute one minus sine squared X. So we'll have 14 multiplied by the quantity of one minus sine squared X. Then bring down the rest plus three sine X minus 12 equals zero. Now distributing that 14, we'll have 14 minus sine squared X plus three sine X minus 12 equals zero. We'll do a couple things here. We'll combine this 14 and the minus 12. And then I want to divide everything by negative one so that we won't have a negative in front of our sine squared X. So this will be a positive sine squared X and then switching signs minus three sine X and then we have a negative 14 plus 12, which is going to be minus two all equal to zero. Now, we can use substitution for sine X. We can substitute any variable we can use U. So this will be 14, not sine squared X but U squared minus three, not sine X but U minus two equals zero. And this one, we can factor, we set up two open sets of parentheses, set those equal to zero. And I'm going to factor this very quickly if you need a refresher on factoring, definitely check out previous lesson videos, but this factors out to be two U minus one and one set of parentheses. And that's multiplied by the quantity of seven U plus two. Now, we can set both of these factors equal to zero and solve for you. So setting the first one, we have two U minus one equals zero, we're going to add one to both sides and we have two U equals a positive one dividing both sides by two, we get U equals a positive one half and that's one of them. Now, let's work on the other. We have seven U plus two equals zero. Subtracting two from both sides, we get seven U equals negative two dividing both sides by seven. And for the other one, we get U equals a negative 2/7. Now that's our U and then we want to go back in and rewrite this instead of you, it will be the sign of X. So we have the sign of X equals a positive one half and the sign of X equals a negative 2/7. So one of these is what right from our tables that we recall from previous lessons, we know that the sign of X is equal to one half at pi divided by six. So we can write X equals pi divided by six. Now we recall from previous lessons if we draw a really quick coordinate plane, recalling our quadrants, the first quadrant is up into the right second quadrant up into the left third quadrant down and to the left and fourth quadrant down into the right of the origin pi divided by six is in the first quadrant. But we know that sign is positive. So we were taking the sign of X to be a positive one half. Well, it's positive in both the first and the second quadrant. So we've got another solution here for X in quadrant two, we can use pi divided by six as our reference angle because that's in the first quadrant. And to find the value in the second quadrant, we're going to take pi minus our reference angle, the pi divided by six and that equals five pi divided by six. So there is another solution that's our quadrant two solution. OK. Those are the two positive ones. Now our other um solution for X, we've got the sign of X equals a negative 2/7. So here it's negative and this one isn't a familiar one right on the unit circle or anything like that. So this one, we get to use the approximations accurate to four significant figures. So when we don't know exactly what it is from the unit circle, we can use the inverse sign. So going to our calculators, we plug in the inverse sign of our negative 2/7 and negative two sevens is within our domain for sign from negative or for inverse sign from negative one to positive one. So this will work, we plug that into the calculator and we'll get a negative 0.28975. And I'm rounding that. Well, we know that the angle negative 0.28975 is in the fourth quadrant. But we do have this restriction that we need to be within the interval from 0 to 2 pi and that's below zero. So we need to get up above zero. So we're going to add two pi to this to find the co terminal angle for it. And if we add two pi then we get 5.9934 we want four significant figures. Well, that's going to be the first four numbers. So the three is the fourth and it looks to the four and it just stays at three. So this one will be 5.993 and that is our fourth quadrant angle here. And it's remember sine X equals negative two sevens. It is negative in the fourth quadrant, but it's also negative in the third quadrant. So we need to figure out where that is in quadrant three. So what we can do, we need to find our reference angle and we recall from previous lessons that our reference angle is going to be the absolute value of our original because here it was negative. So if we take the absolute value of the inverse sign of negative 2/ we're going to get a positive 0.28975. So that's in the first quadrant. Now to figure out what the third quadrant angle is going to be, we just add pie. And so that is going to get us 3.41 or excuse me, 3.4313 to 4 significant figures. The fourth number here is the significant one. That's the one it looks to the three, it's just going to stay a one. So our third quadrant angle is 3.431. And those are the four solutions for X. For this one, we look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

In Exercises 121–126, solve each equation on the interval [0, 2𝝅). 10 cos² x + 3 sin x - 9 = 0

Key Concepts

Trigonometric Identities

Solving Quadratic Equations in Trigonometric Form

Interval Restriction and Solution Verification

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