In Exercises 63–84, use an identity to solve each equation on the...

Question

In Exercises 63–84, use an identity to solve each equation on the interval [0, 2𝝅).

2 cos² x + sin x - 1 = 0

Accepted Answer

Welcome back everyone. In this problem, we want to solve the trigonometric equation two multiplied by the square of sine X plus cosine X minus one equals zero on the interval. Open bracket. +02 pi closed parentheses. With the help of an identity for our answer choices. A is two thirds of pie and four thirds of pie, B is zero and two thirds of pie, C is zero and four thirds of pie and D is 02 3rd of pie and four thirds of pie. No, we want to solve it with the help of an identity. So can you recall any identity that includes either the square of sine X or cosine X? Well, recall that the square of sine X equals one minus the square of cosine X. Another way to write the Pythagorean identity. So we can use that identity to replace the square of sine X in our equation and thus sulfur X using the cosine function. So let's go ahead and do that because no, that means we can rewrite our equation as two multiplied by one minus the square of cosine X plus cosine X minus one equals zero. If we distribute 22, multiplied by one equals two and multiplied by the square of cosine X equals two, multiplied by the square of cosine X plus cosine X minus one equals zero. We can combine the like terms two and negative one. Now that leaves us with negative two multiplied by the square of cosine X plus cosine X plus one equals zero. No, our squared cosine X term is negative or squared term is negative. So, and we can make it positive by multiplying through by negative one. So multiplayer the equation by negative one. And that's going to help us to factorize. It means that we're going to multiply negative two cosine squared X by negative one that makes a positive two cosine squared X cosine X multiplied by a negative one equals negative cosine X and one multiplied by a negative one equals negative one. And on the right hand side zero, multiplied by a negative one still equals zero. No, we can go ahead and factorize and by factorizing, you'll notice that we end up with two multiplied by the cosine of X plus one multiplied by the cosine of X minus one. And that all equals zero. Now given that this equals zero, given the product equals zero, that means one of the terms must be equal to zero. So this tells us then that either two multiplied by the cosine of X plus one equals zero or the core sign of X, the cosine of X minus one equals zero. So now we can go ahead and solve for both. Let's start with two multiplied by the cosine of X plus one equals zero. That means two multiplied by the cosine of X equals negative one. And thus the core sign of X would be equal to negative one divided by two or negative a half. So now let's think about it. What are the possible values of X to make the cosine negative a half? Let's use our unit circle to help us. Now recall that on the unit circle, let me bring it a little more in, in focus, recall that on our unit circle where we have our Y axis and our X axis, the coordinates on our unit circle are written in the form the cosine of theta and the sign of the, where the cosine of theta is the X value and the sign of theta is the Y value. Now for the cosine of X or the cosine of our value being equal to negative or half where X is our angle. Then that tells us that the X value at that point would be negative or half. So let's say it would be right there. And at that point, you notice that we have two possible values for the cosine for the angle for the cosine one in the second quadrant and one in the third quadrant. Now notice that at both of those points recall that it would have been either two thirds of pi and this in the second quadrant, that is my apologies. And in the third quadrant, it would be the same distance away from the X axis. So if it's two thirds of pi in the second quadrant, it would be a third of pi distance away in the third quadrant. So that would be four thirds of pi. So therefore, this tells us that X equals two thirds of pi and it also equals four thirds of pi when the cosine of X equals negative a half, we have one more to solve. And that is when the cosine of X minus one equals zero. If that's true, that means the cosine of X equals one. And let's think what value of X allows the cosine of X to be equal to one. Well, again, if we draw for our unit circle, then on our unit circle, we want the value of X when the cosine equals one. So that means the X value of the cosine would be equal to one. And at that point, our angle would be equal to zero. So therefore, X or angle would be equal to zero degrees. And that's the only possible value for the cosine when the cosine of X equals one. So that tells us then that X could be equal to zero, could be equal to two thirds of pi or it could be equal to four thirds of pi. Therefore, when we look back on our answer choices that would have been answer choice. D thanks a lot for watching everyone. I hope this video helped.

In Exercises 63–84, use an identity to solve each equation on the interval [0, 2𝝅). 2 cos² x + sin x - 1 = 0

Key Concepts

Trigonometric Identities

Pythagorean Identity

Solving Trigonometric Equations

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