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.

sec² x + 3 sec x + 2 = 0

Accepted Answer

Hello, today we're gonna be solving the following trigonometric equation. So what we are given is seeking squared of X plus second of X minus two is equal to zero. Now, we're first going to need to simplify the left hand side of the equation. Notice that the left hand side of the equation roughly resembles a polynomial. If we were to use a substitution, let's say Y is equal to C and of X, we can simplify the equation as Y squared plus Y minus two is equal to zero. Currently, this is in the form of a factor trinomial. And the left hand side of the equation will factor to give us the values of Y minus one multiplied by Y plus two is equal to zero. We can then set both factors of Y equal to zero, leaving us with Y minus one is equal to zero and Y plus two is equal to zero. Next, let's go ahead and solve for Y in both of the equations. On the left hand side, if we add one to both sides of the equation, we get Y is equal to positive one. And if we were sub to subtract two from both sides of the equation. On the right hand side, we are left with Y is equal to negative two. Now recall that we stated that Y is equal to seeking of X using res substituting seekins into the equations will leave us with seeking of X is equal to one and seeking of X is equal to negative two. Now, if you're unsure of how to find these angles, you can always rewrite seeking in terms of cosine. Recall that seeking of X is the inverse of cosine or in other words, seeking of X is equal to one over cosine of X. So we can replace seeking of X with one over cosine of X and rewrite both of the equations as one over cosine of X is equal to one and one over cosine of X is equal to negative two, we will then need to solve for cosine of X. In both of the equations. On the left hand side, we can multiply cosine of X to both sides of the equation. And on the right hand side, we can multiply both sides by cosine of X. This will allow us to rewrite the equations as one is equal to cosine of X and one is equal to negative two cosine of X. And on the right hand side, if we were to divide to divide both sides of the equation by negative two, what we are left with is cosine of X is equal to negative 1/2. So now, in order to solve for X, we need to ask ourselves, what values of X can we plug in a cosine to give us the value of one? And what values of X can we plug into cosine to give us the value of negative 1/2? While in order to solve this, we'll need to use the unit circle. Now recall that any terminal point on the unit circle is represented as cosine commas sign cosine represents any X value on the unit circle. And sine represents any Y value on the unit circle. Now, starting with the first equation, what angles have cosine of X equal to one? Or in other words, what angles have a terminal point of one in the X value? Well, at the angle zero and two pi we have the terminal point of one comma zero at both of these angles cosine is equal to one. And recall that we want all the values of X within the region of to 2 pi including two pi. So the values of X for the first equation are going to be X is equal to zero and X is equal to two pi. Now we want to ask ourselves for the second equation, what terminal points have negative 1/2 in the X values of the unit circle? Well, there are two total angles. The first angle lies within the second quadrant at two pi over three, the terminal point at this angle is negative 1/ comma square root 3/2. And the second angle lies within the third quadrant at four pi over three. And this terminal point is given to us as negative 1/ comma negative square root 3/2. So what this means that for the second equation, the values of X that we can plug in a cosine are going to be X is equal to two pi over three and X is equal to four pi over three. So in total, we have four solutions of X that make the original equation true X is going to equal to 02 pi over 34 pi over three and two pie. And with that being said, the answer to this problem is going to be B 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. sec² x + 3 sec x + 2 = 0

Key Concepts

Secant Function

Trigonometric Identities

Interval Notation

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