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

Question

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

Accepted Answer

Hello, today we're going to be solving the following trigonometric equation on the interval from 0 to 2 pi. So what we are given is tangent of X minus seeking of X is equal to negative one. Now, in order to solve for X, we're first going to need to find a way to simplify the given equation. Well note that we have a tangent term and a second term given to us, we do have one trigonometric identity that uses tangent. And second that trigonometric identity is given to us as tangent squared X plus one is equal to C squared of X. And if we were to subtract both sides of this identity by one and subtract both sides by C and squared, we can rewrite this identity as tangent squared of X minus C and squared of X is equal to negative one. So what we'll need to do is we'll first need to find a way to get tangent squared X minus C and squared X and use this identity to simplify the given equation. While the first step in doing that is we can first add seeking of X to both sides of the equation. Doing so allows us to rewrite this as tangent of X is equal to negative one plus seeking of X. Now note that the identity has tangent squared and C can squared. In order to get squared values in our equation, we're going to have to square both sides of the given equation. On the left hand side, tangent X squared gives us tangent squared of X. And on the right hand side, if we were to foil negative one plus seek in X, we can rewrite the right hand side as one minus two seeking of X plus C N squared of X. Now note that the identity has a difference of tangent squared X and C can squared X. We want to get both the tangent squared and C can square terms on one side of the equation. And in order to do that, we can subtract both sides of the equation by C and squared of X, we can then rewrite the left hand side as tangent squared of X minus C squared of X is equal to one minus two second of X. Now, on the left hand side, we have tangent squared X minus C can squared X and going back to the trigonometric identity, the identity states that tangent squared X minus C can squared X is equal to negative one. So we can use the identity to rewrite the left hand side of the equation as negative one. And the simplified form of the equation is going to be negative one is equal to one minus two second of X. Now, what we wanna do is you want to isolate the seek in term in order to do this, we can subtract one to both sides of the equation allowing us to rewrite it as negative two is equal to negative two sequent of X. And if we were to divide to divide both sides of the equation by negative two, what we are left with is seeking of X is equal to positive one. So what this means is what angle of X can we enter into sequent to give us the value of one? Well, if you're ever unsure of how to use a, see in term, we can always rewrite seeking in terms of sign and cosine recall that the trigonometric identity for second is the inverse of cosine or in other words, sin of X is equal to one over cosine of X. So if we were to use this trigonometric identity, we can rewrite the left hand side as one over cosine of X is equal to one, we then want to solve for cosine of X. And we can do that by multiplying both sides of the equation by cosine of X. This will eliminate cosine from the denominator leaving us with one is equal to cosine of X. Now what this simplified form states is what angle of cosine is going to give us the value of one within the range of 0 to 2 pi not including two pi. Well, in order to answer this, we can go ahead and use the unit circle. Now recall that any terminal point on the unit circle is defined by cosine comma son. What this means is that cosine represents any X value on the unit circle or on the terminal points of the unit circle. And sine represents the Y values of the terminal points of the unit circle. Now, at the angle zero degrees, we have the terminal 00.1 comma zero. And this statement is true for the angle two pi. Now again, the values for XX can only lie within the first rotation of the unit circle, not including the value of two pi. And so what this means is that if we were to choose the value of zero for X cosine of zero is going to give us one. So that means the value of X is going to equal to zero. And with that being said, the answer to this problem is going to be D. 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 63–84, use an identity to solve each equation on the interval [0, 2𝝅). tan x + sec x = 1

Key Concepts

Trigonometric Identities

Solving Trigonometric Equations

Domain and Interval Considerations

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