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. 7 cos x = 4 - 2 sin² 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 or providing approximations accurate to four significant figures. Our equation is two minus nine sine squared X equals five cosine X. Our answer choices are answer choice A X equals the solution set 2.274 radiance and 4. radiance. Answer choice B X equals the solution set negative 0.7034 radiance and 2. radiance. Answer choice C X equals the solution set negative 0.5742 radians and answer choice D no solution. All right. So looking at our equation here, we can do a few things, we can move this around so that everything is on the left hand side of the equal sign and then on the right hand side, it's equal to zero. So we'll subtract five cosine X from both sides. And so on the left, we'll have two minus nine sine squared X minus five cosine X equals zero. And now we've got a sign squared and a cosine, we recall from previous lessons that the sine squared of X equals one minus the cosine squared of X. So we can use substitution here for the sine squared of X, we can substitute one minus the cosine squared of X. So this will be two minus nine multiplied by the quantity of no longer sine squared X. But now one minus cosine squared X minus five cosine X equals zero. So now everything at least is in cosine. So let's distribute this negative nine to the one into the negative cosine squared X. So we'll have two minus nine plus nine cosine squared X minus five cosine X equals zero. Now let's combine, combine like terms and put these in descending order. So we'll start with the highest degree and then going down in degrees. So we'll start with the nine cosine squared X minus five cosine X and then it's just two minus nine, which is minus seven equals zero. And now we've got this lovely quadratic and we can substitute for our cosine X. We can choose any variable. I'm going to choose you. So we'll have nine multiplied, not by cosine squared X, but by U squared minus five multiplied, not by cosine X, but by U minus seven equals zero. And here we've got this quadratic but it's, it's really not that lovely because for this one, we get to use the good old friends, the quadratic formula. So we recall from previous lessons that that is negative B plus or minus the square rut of B squared minus four AC all divided by two A recognizing that our A here in front of the U squared is nine, our B in front of our U is negative five and our C in front of, well, the constant term is negative seven. So if we plug all of that in, we'll get a solution. Now for the sake of time, I am not going to go through all of the steps for the quadratic formula. If you need a refresher on the quadratic formula, definitely check out previous lesson videos. But for the sake of time, we will get a solution when we plug in for A B and C, we'll get five plus or minus the square root of 277 all divided by 18. And so those are our two solutions. That's what U is equal to. But recalling that we're not talking about you here. We're talking about the cosine of X. So we can set the cosine of X equal to five plus the square rut of 277 all divided by 18. And we can set the cosine of X equal to five minus the square rut of all divided by 18. So now we've got the cosine of X equal to something we can use our calculators to plug in to solve for this using the inverse cosine. So X will equal the inverse cosine as five plus the square root of 277 all divided by 18. But when you plug that in to your calculator, you're going to get a domain error for that one. Why? Well, because five plus the square root of 277 divided by 18 is approximately equal to 1.20241. And the domain for our inverse cosine is from negative 1 to 1, 1.20241 is outside of that domain. So we don't need to worry about that part of it. But we can then go forth and put in, we know that X now equals the inverse cosine of five minus the square rut of 277 all divided by 18. Because that one, the five minus the square root of 277 divided by 18, that one's equal to approximately negative 0.64685. So that's within our domain. OK? Now, when we plug that into the calculator, we will get a solution for X of 2.2742. And if you don't get that, be sure that you are in the correct mode, make sure you're in radiance. So 2.2742, if we think for a moment about a coordinate plane, we've got our vertical y axis, our horizontal X axis, they come together at the origin in the middle. And if we're in radiance, the negative part of the x axis, that's pi and pi is about 3. half of that, the positive part of the Y axis is about 1.5. So if we have 2.2742, that's going to be in quadrant two. And that makes sense because that is where cosine is negative, but where else is cosine negative? It's also negative in quadrant three. So we've got another solution for this. We have our quadrant two solution of X equals 2.2742. But we need our quadrant three solution. So in order to get that, there's a couple of ways going about doing it, we can figure out our quadrant one reference angle. And in order to get that, we would take pi minus our quadrant two angle which to be specific in the calculator is the cosine the, excuse me, the inverse cosine of five minus the square of 277 all divided by 18. And then you take that solution and you add pi in order to get our quadrant three solution. And when you do that, you'll get an X value of 4.89. Now, the only thing left is that we need to have these accurate 24 significant figures. So looking at X equals 2.2742. The fourth digit is the four, the four looks to the two, it's not going to move. So we have an X value of 2.274. And for our quadrant three angle 4.8, the eight is the fourth significant figure, significant digit. It looks to the nine and that one rounds up. So that's going to be 4.9 and that is our solution set. Now, we look at our answer choices and that matches with answer choice. A 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. 7 cos x = 4 - 2 sin² x

Key Concepts

Trigonometric Identities

Solving Trigonometric Equations

Interval and Exact vs Approximate Solutions

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