In Exercises 127–130, solve each equation on the interval [0, 2𝝅...

Question

In Exercises 127–130, solve each equation on the interval [0, 2𝝅) by first rewriting the equation in terms of sines or cosines.

csc² x + csc x - 2 = 0

Accepted Answer

Hello, today we're going to be solving the following trigonometric equation. So what we are given is coin squared of X plus three. Coking of X plus two is equal to zero. Notice that the left hand side of this equation roughly resembles a polynomial. We can represent the left-hand side as a polynomial by applying a substitution. If we allow Y to equal to coking of X, we can represent the left-hand side as Y squared plus three, Y plus two is equal to zero. The left hand side is in the form of a factor trinomial and this trinomial will factor to give us Y plus one multiplied by Y plus two is equal to zero. We can then set the factors of Y equal to zero and end up with two separate equations. Y plus one is equal to zero and Y plus two is equal to zero. Now, if we were to res substitute Y with coin of X, we can rewrite the given equations as coking of X plus one is equal to zero and coin of X plus two is equal to zero. We'll need to solve for coin of X in both of the equations. And on the left hand side, we can subtract both equations by one to give us coin of X is equal to negative one. And on the right hand side, if we subtract two to both sides of the equation, we end up with king of X is equal to negative two. So in order to further simplify this equation, we can rewrite coking of X in terms of sine recall that coin of X is defined as one over sign. If we were to apply this identity to both of the equations, we can rewrite the equations as one over sine of X is equal to negative one and one over sign of X is equal to negative two. If we were to multiply both sides of both of the equations by sin of X, we can rewrite both equations as one is equal to negative sign of X and one is equal to negative two sign of X. On the left hand side, if we divide both sides of the equation by negative one, what we are left with is sign of X is equal to negative one. And on the right hand side, if we divide both sides of the equation by negative two, we are left with sign of X is equal to negative 1/2. So this is going to be the simplified form of both of the equations. And now the question becomes what angles of X can we plug in a sign to give us the value of negative one at one angle of X. Can we plug in a sign to give us the value of negative 1/2? In order to answer this question, we're going to need to use a unit circle. Now recall that any terminal point on a unit circle is defined as cosine comma sign here cosine represents any X value on the unit circle and sine represents any Y value on the unit circle. Let's go ahead and start by solving for sin of X is equal to negative one. What this means is what angle within the range of 0 to 2? Pi has negative one in the Y value of its terminal point? Well, at the angle of three pi over two, the terminal point at this angle is given to us as zero comma negative one. This is the only angle between zero and two pi that will give us the value of negative one if we were to plug it into sine. And what this means is that for sine of X is equal to negative one, X is going to equal to three pi over two. Now let's solve the second equation. What angles of X can we plug in a sign that will give us the value of negative 1/2. Well, there are two angles that answer this question. The first angle lies in the third quadrant of the unit circle at seven pi over six. And the terminal point at this given angle is negative square root 3/2 comma negative 1/2. The Y value at this angle is negative 1/2. And if we were to plug in seven pi over six into sine, we get the value of negative 1/2. So that means X is going to equal to seven pi over six, we also have another angle that gives us the value of negative 1/2. That value or that angle lies in the fourth quadrant at pi over six. The terminal point at this angle is square root 3/2 comma negative 1/2. And the Y value at this angle is negative 1/2. So that means the second value of X that we can plug in a sign is going to be 11 pi over six. So in total, there are three solutions to X within the domain of 0 to 2 pi X is equal to seven pi over six, three pi over two and 11 pie over six. These are going to be the values of X that answer the, the given equation. And with that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 127–130, solve each equation on the interval [0, 2𝝅) by first rewriting the equation in terms of sines or cosines. csc² x + csc x - 2 = 0

Key Concepts

Cosecant Function

Trigonometric Identities

Interval Notation

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