In Exercises 53–62, solve each equation on the interval [0, 2𝝅)....

Question

In Exercises 53–62, solve each equation on the interval [0, 2𝝅).                   tan² x cos x = tan² x

Accepted Answer

Hey, everyone in this problem, we're given the trigonometric equation cotangent squared of X multiplied by sin of X is equal to cotangent squared of X. And we're asked to solve in the interval from 0 to 2 pi excluding the end 0.2 pi. And to give the answer, an exact gradient measure, we have four answer choices. Option A is that X is equal to two. Pi option BX is equal to three pi divided by two. Option CX is the set containing pi and pi divided by two. And option DX is the set containing pi divided by two and three pi divided by two. So let's get started by writing out our question. We have code changes squared of X multiplied by sin of X is equal to cotangent squared of X. We're gonna treat this just like any other equation. OK. We don't want to get scared or overwhelmed that there are trigonometric functions here. Let's get everything to one side. We'll have zero on the other side and we can see what we can do with this. So moving our cotangent squared of X to the left hand side, we get that cotangent squared of X multiplied by sin of X minus code change is squared of X is equal to zero. Now, on the left hand side, we can fact we have this cotangent squared of X in each term. So we factor that oh we have cotangent squared of X multiplied by sin X minus one is equal to zero. Now the left hand side is back. So if either of these factors is equal to zero, the entire left hand side will be equal to zero and the equation will be satisfied. OK. So let's look at co change at square of X first, if we have cotangent squared of X is equal to zero, that tells us that cotangent of X must be equal to zero. So cotangent is one divided by tangent of X. We called the tangent of X is sine divided by cosine. And so our cotangent is equivalent to cosine of X. By now, we have a restriction here. If sine is equal to zero, if sine is equal to zero, then the function will be undefined. So we wanna avoid any point where sine is zero. OK. So that's gonna be zero or pi within that 0 to 2 pi interval excluding two pi. But the only way this can be equal to zero co X divided by sin X is if cosine of X is equal to zero. OK. So we have that cosine of X is equal to zero. Well, when does cosine of X equal to zero. We know that we can use our unit circle or we can use our graph of cosine. Can I recall what the graph of cosine looks like? Roughly starts at one decreases down to the minimum and backup. And we know where these zeros will occur. They occur at pi divided by two and at 35 divided by two. OK. So we know that we have these two solutions pi divided by two and three pi divided by two. Now let's look at our second factor and see if there are any other solutions. OK. So we have five, X minus one is gonna be equal to zero. Tells us that sin X is equal to one. And if we do the same thing, if we draw our graph of time starts at zero, increases to one before decreasing and coming down to its minimum and back to the origin at two, we can see that this is gonna occur at pi divided by two as well. We have X is equal to pi divided by two as a solution here, but that's already included in the solution we found. OK. So we don't need to include it twice. So what we found is that there are two X values that satisfy this equation. We have X is equal to pi divided by two or three pi divided by two, which corresponds with answer choice. D. Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 53–62, solve each equation on the interval [0, 2𝝅). tan² x cos x = tan² x

Key Concepts

Trigonometric Identities

Solving Trigonometric Equations

Interval Notation

Watch next