In Exercises 54–67, solve each equation on the interval [0, 2𝝅)....

Question

In Exercises 54–67, solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places.

2 sin² x + sin 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 five sine squared X plus two sine X minus four equals zero. Our answer choices are answer choice A X equals the solution set 0.2433 and 7.898. Answer choice BX equals the solution set 0.7988 and 2.343. Answer choice CX equals the solution set 42.63 and 45.77. And answer choice DX equals the solution set 45.75 and 46.32. All right. So we've got this nice little quadratic here. And what we can do is we can use substitutions so that this reads a little bit more intuitively. We can take sine X and substitute that with you. And so we'll have five multiplied not by sine squared X. But now by U squared plus two multiplied not by sine X, but by U then minus four equals zero. And it is a lovely little quadratic but it doesn't factor very easily. So we get to bring out our friend the quadratic formula. Recalling from previous lessons, the quadratic formula is negative B plus or minus the square rot of B squared minus four AC all divided by two A. And here our A term is five in front of the U squared. Our B term is two in front of the U. And our C term is negative four, if you plug all of those in and then go through the whole process to simplify. If you need a refresher on the quadratic formula, definitely check out previous lesson videos. But for the sake of time, I'm just going to go with the solution for this one which is going to be negative one plus or minus the square root of 21 all divided by five. So we have two answers here that we, we know that our U is equal to, but we also know that U is the sign of X. So we can say the sign of X is equal to. And for the first solution, we have negative one plus the square ret of 21 all divided by five. And for the second one, we know that the sign of X is equal to negative one minus the square rat of 21 all divided by five. And now noting what those would be with decimal places. So these aren't exact answers, but it helps to know approximately what that is. For the first one, we're negative one plus the square of 21 all divided by five. That's approximately 0.716515. And for the second one where we have negative one minus the square rot of 21 all divided by five, that is equal to approximately negative 1.11652. So setting that aside for a moment, we know that the sign of X is equal to those two things. But what about X? What is X equal to? We recall from previous lessons here, we use our calculators and we use the inverse sign function. So we know that for the first one, X equals the inverse sign of that number, that negative one plus the square of 21 all divided by five. And we put that into a calculator and we will get 0.79879. And we know that if you think about for a moment, OK. We know that pi is the straight line that's 180 degrees. If we've got a little coordinate plane here, pi is the left side of the X axis heading toward negative infinity. And so if we've got a value in radiance, make sure you have your calculator in radiance that 0.79879 for the angle. Well, pie is about 3.14 and half of that is about 1.5. So we know that 0.7987 is in the first quadrant and sign is positive in both the first and the second quadrant. So we need to figure out what this angle would be in the second quadrant as well. Well, we know finding a second quadrant angle, we take pi minus our reference angle. And in this case, our reference angle is the one from the first quadrant. It's that 0.79879. But you can even put in to be more accurate. You can put in the um pi minus the inverse sign of negative one plus the square of 21 all divided by five. And then you will get your very accurate quadrant two answer for X which is going to be 2.3427. So those are two values we saw in the instructions that they need to be to four significant figures. So for the first one, the fourth value after the decimal point, the seven is the fourth significant figure that zero is not significant. So this is going to be 0.79 eight and then the seven will round up because of the nine. So 80.7988 is one solution here. And for 2.3427 that two, the fourth digit is the fourth significant digit. It looks to the seven and it will round up. So 2.343 is our other solution here for X. And now we've got this second equation to take a look at the sign of X equals negative one minus the square root of 21 all divided by five. So if you were to plug in to your calculator solving for X, we have the inverse sign of that negative one minus the square root of 21 all divided by five, you will get a domain error. And that is because for the domain of our inverse sign X needs to be in between negative one and positive one. And here we know that negative one minus the square root of 21 all divided by five is equal to approximately negative 1.11652 which is outside of our domain. So that's why we only get these two solutions. Here, 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 54–67, solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places. 2 sin² x + sin x - 2 = 0

Key Concepts

Quadratic Equations in Trigonometry

Trigonometric Functions and Their Values

Interval Notation and Solutions

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