In Exercises 50–53, find all solutions of each equation.

1
cos x...

Question

In Exercises 50–53, find all solutions of each equation.

1
cos x = ﹣-----
2

Accepted Answer

Hello, today we're going to be solving the following trigonometric equation and giving the answer in radiance. So what we are given is cosine of A is equal to negative square root 3/2. So currently the equation is given to us in a simplified form. And the way to read this is what angles of a can we plug into cosine that will give us the value of negative square root 3/2. Well, in order to answer this question, we're going to need to use a unit circle. Now recall that any terminal point that lies on the unit circle is defined as cosine comma sine. This means that cosine represents any X value on the unit circle and sine represents any Y value on the unit circle. So the question is what terminal points have the value of negative square root 3/2 in the X value of the unit circle. While there are two angles that give us this value, the first angle lies in quadrant two at five pi over six. And the terminal point of this angle is negative square roots 3/ comma one half the second angle that gives us, this value lies in the third quadrant of the unit circle at the angle seven pi over six. And the terminal point for this angle is negative square root 3/ common negative 1/2. So both of these angles have negative square over 3/2 in the X value of the terminal point. And that means the angles that we can plug in a cosine to give us this value is going to be A, is equal to five pi over six and A is equal to seven pi over six. Now, there's one thing to specify about this question and that is the fact that we have no specified range of a within the rotation of the unit circle. In other problems, we are normally told that we want the angle to be within the range of 0 to pi. But in this case, we are not given this condition. So we're going to have to account for the fact that we have an infinite amount of angles that represent the same terminal point. So how do we represent an infinite amount of angles with the given angles? Well, we can go ahead and add the period for cosine to five pi over six and seven pi over six. Now recall that the period for cosine is given to us as two pi. So every rotation of the unit circle for cosine will be two pi and that means A is going to equal to five pi over six plus two pi and A is equal to seven pi over six plus two pi. Now, we still need a variable that represents the amount of rotations that we can find this value at. And we can represent the integer as N and is going to represent the amount of rotations that we make around the unit circle that gives us the same value. And with that being said, the answer to this problem is going to be A is equal to five pi over six plus two pi N and A is equal to seven pi over six plus two pi N. And the correct choice for 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 50–53, find all solutions of each equation. 1 cos x = ﹣----- 2

Key Concepts

Cosine Function

Inverse Trigonometric Functions

General Solutions of Trigonometric Equations

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