In Exercises 121–126, solve each equation on the interval [0, 2𝝅...

Question

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

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 two pi employing exact values whenever feasible and providing approximations accurate to four significant figures. Our equation is two cosine squared X equals seven cosine squared X minus four sine X. Our answer choices are answer choice. A no solution answer choice B X equals the solution set 0.7437 and 2.398. Answer choice C X equals the solution set pi divided by six and five pi divided by six. And answer choice D X equals the solution set 0.2573 and 6.26. All right. So a few things we notice about our equation here, we've got a couple of like terms. We have this two cosine squared ax. And then on the other side of the equal sign seven cosine squared X and then our third term is a sine X. So those are not like terms. What we can do is we can subtract the two cosine squared acts from both sides to bring our like terms together. And then I'm just going to write this with the right hand side. Now on the left, so seven cosine squared X minus two cosine squared X is five cosine squared X and then our minus four sine X equals. And now writing what was on the left hand side, on the right hand side is now zero. All right. So we've got a cosine squared and a sign squared. But we remember from our previous lessons on identities that cosine squared X is equal to one minus sine squared X. So we can substitute for cosine squared X A one minus sine squared X. So we'll have five multiplied by the quantity of, instead of cosine squared X, we'll have one minus sine squared X then bring over minus four sine X equals zero. Now, we can distribute that five. So we'll have five multiplied by one is five minus five, multiplied by sine squared X is five sine squared X minus four sine X equals zero. And we've got a quadratic here. We can put these in order. So start with this negative five sine squared X minus four sine X plus five equals zero. And now the last thing here, I wanna make sure we get rid of this negative in front of our five sine squared X. So I'm dividing everything by negative one and we will have a positive five sine squared X and then that switches to A plus four sine X and switches to A minus five equals zero. Now, we can use substitution again for our sine X that can be A U. And so we'll have five multiplied not by sine squared X, but five U squared and then plus four, not multiplied by sine X but four U and then minus five equals zero. We've got this lovely little quadratic but it's really not so lovely. We can't factor it very easily. So noticing that our, a term here is five, our B term here is four. Our C term is negative five. We can utilize the quadratic formula to solve here for you. And the quadratic formula we recall we recall from previous lessons is negative B plus or minus the square root of B squared minus four AC inside the radical and everything is divided by two A. Now, if you need a refresher on the quadratic formula, definitely check out previous lesson videos. For the sake of time, I'm going to do this very quickly. Substituting for A's B's and C's, this is going to equal for the final product of this. A negative two plus or minus the square of 29 all divided by five. So our U is equal to that. And so then substituting that in for you, we are saying that sign of X equals two different things. One is negative two plus the square of 29 all divided by five. And the other is that the sign of X is equal to negative two minus the square of 29 all divided by five. And it could be helpful to know what that is in a decimal even though it's not as exact. So for the negative two plus the squared of 29 all divided by five that equals 0.67703. And then it keeps going, but leaving it there. And for the negative two minus the square root of 29 divided by five, that is equal to a negative 1.477 oh three. So we've got the sign of X equal to two different things. Now, how do we solve for X? Will we recall from previous lessons that we can take the inverse sign of our two solutions? And that will give us X so X equals the inverse sign of negative two plus the square of 29 all divided by five and X equals the inverse sign of negative two minus the square of all divided by five. And so now it's just a matter of solving for these. So let's start with the X equals the inverse sign of negative two plus the squared of 29 divided by five. We plug that into a calculator and you wanna make sure that you're in the correct mode, you wanna make sure that you are in radiance and you will get that X equals 0.74372. But we want to be to four significant figures. So the zero isn't significant 7437, that seven just stays a seven because the two after it, so 0.7437 is one of our solutions for X. Now we know that 0.7437 is in the first quadrant. So remember we're talking about radiance and anything in the first two quadrants, if we draw a quick sketch of a coordinate plane, the first quadrant, the upper right and then the second quadrant to the left of that, we know that the negative part of the X axis, that's pi which is about 3.14 and so half of that is just over 1.5. And so our 0.7437, that's in the first quadrant. And we've got sine which is positive as this number is in both the first and the second quadrant. So we can treat this as a reference angle for our value in the second quadrant. And we recall from previous lessons that we would find the value in the second quadrant by taking pi minus our reference angle. So our angle, our quadrant one angle which is 0.7437. And you can put in the exact, you could put pi minus negative two plus the square root of 29 divided by five. And you are going to get 2.3978. But rounding to four significant figures, that's just going to be the 2.3977 looks to eight and that's going to round up and be 2.398. And that's our quadrant two value for Sine because Sine's positive and quadrants one and two. So we've got two answers so far. Now we need to take another look at our X equals the sign or excuse me, the inverse sign of negative two minus the square root of 29 divided by five. But we see that that is equal to in decimal form, a negative 1.47703. And the domain for the inverse sign is from negative one to one. And that goes further toward negative infinity, the negative one that's outside of our domain. So when you plug that into your calculator, you'll get a domain error. And so we can disregard anything for that second one, for the negative two minus square root of 29 all divided by five. And that's how we get our solution set of 0.7437 and 2.398. We look at our answer choices and that matches with answer choice. B well done. We'll catch you on the next one.

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

Key Concepts

Trigonometric Identities

Solving Trigonometric Equations

Interval Notation

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