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).

(tan x - 1) (cos x + 1) = 0

Accepted Answer

Hey, everyone in this problem, we have the trigonometric equation copes X minus one multiplied by 10 X plus the squared of three is equal to zero or we will solve in the interval from 0 to 2 pi including the end points. And to give the answer, an exact radiant measure, we have four answer choices here. Option A and B are both sets containing four possible X values. OK? And they just differ in what those values are. Option C is a set containing three values for X and option D is that there is no solution. Let's start by writing out the equation we were given. So we have a cosine of X minus one. We applied by tangent of X plus the square root of three is equal to zero. OK. And we don't wanna get overwhelmed by the fact that we have trigonometric functions inside of this equation. We want to treat this like any other equation we deal with. OK. So right now we have two factors multiplied together on one side, we have zero on the other side. So we know that if either of these factors on the left hand side is equal to zero, then the entire thing will be equal to zero. And that will satisfy our equation. So we can deal with each of these factors one at a time, starting with the first one, we have the zone of X minus one is equal to zero. We can add one to both sides. We get that cosine of X is equal to one. And I don't wanna think about when this is true. Now, you can think about this in terms of your unit circle, I like to think about it in terms of the graph of cosine. And so we drew draw a rough sketch of the graph of cosine. Recall that cosine starts at one when X is equal to zero, it comes down to its minimum value of negative one and then returns back to positive one after one complete cycle. OK? Or two P radians. So we can see that cosine of X is gonna equal one at X equals zero or X equals two pi both of those are in our interval. OK. And we don't have to worry about anything further than that to the left or the right because that's outside of our interval. So we get these two solutions X is equal to zero or two pi we are gonna circle these in green. So we can come back to them. And from our answer traces, we can kind of narrow it down to either option A or C. OK. Both of those contain zero and two pi as options, option B does not contain two pi so we can eliminate that. And option D is that there's no solution, but we found that there is a solution so we can eliminate that one as well. Now I'm gonna work with the second factor that we had, which is tangent of X plus the square root of three is equal to zero. Now, we're gonna treat this the same way we did the cosine equation. We're gonna move our square root of three to the right by subtracting. So we get that the tangent of X is equal to negative the square root of three. No tangent is a negative value. Where is that gonna occur. OK. Well, tangent is going to be negative in quadrants two and quad four right now, let's think about our reference angle and our reference angle is gonna be an angle between zero and pi divided by two. And that relates this tangent to this value of the square of three. OK. We're gonna look just at the positive value for right now if we think about our special triangle, OK. And it's really good to remember these special triangles. So we're called the special triangle where we have side links one it too, which means the other side length is a square root of three. And you can always double check these side length through Pythagorean theorem because this is a right angle triangle. OK. So if we do Pythagorean theorem, one square plus the square root of three squared gives us four, we take the square root and get the hypo. Now we know that across from the smallest side, which is one is gonna be our smallest angle. And so our angle across from one is gonna be pi divided by six for this particular special triangle. And across from pi divided or sorry, across from the square root of three, we get pi divided by three. Now the tangent is gonna be the opposite side divided by the adjacent side. OK. So looking at our special triangle, tangent of pi divided by three is gonna be the squared of three divided by one, which is exactly what we have with the square root of three. And so our reference angle X prime is gonna be pi divided by three. Now we have a reference angle but that doesn't mean that that is our solution. OK. We need to think about where tangent is negative. OK. We have this negative value. We've said it's gonna be negative in Q two and Q four. So let's go ahead and draw out what we have so far. OK. It's gonna be a really helpful way to visualize what we have and to figure out how to actually find these angles. So again, we're talking about Q two, which is the top left and Q four the bottom right. And we're gonna draw this pi divided by three angle. But recall the reference angles are always measured from the nearest x axis. So for pi divided by three and Q two, it's gonna be pi divided by three from the negative X ax sorry ne X axis upwards in the clockwise direction. And in quadrant four, it's gonna be pi divided by three measured from the positive X axis in the clockwise direction as well. So if we wanna find the actual answer to this question, the actual X values for these angles, we need to measure them from the positive X axis in the counterclockwise direction. And that's the convention. So we're gonna go from the positive X axis up, counter clockwise till we hit the solution in quadrant two. And we're gonna call that value X two. And we can see that X two. Well, if we go all the way to the negative X axis, that's gonna form a straight line, that's gonna be pi radiance, but we're stopping pi divided by three short. And so X two is gonna be pi radiance minus pi divided by three radiant which gives us two pi divided by three ratings. Now, for the second solution for this factor, OK, we're gonna do the same thing. We're gonna go from the positive X axis all the way counterclockwise until we hit the solution in quadrant four. And we're gonna call that X four. Now X four, we can see that if we go all the way around this circle, that's two pi radiant and we know that a circle is two P radium. So going all the way around will be two P radians or we're stopping pi divided by three radiant short. So X four is gonna be two pi minus pi divided by three radiant. And this is gonna give us five pi divided by three radiates. So now we found four solutions two from our first factor and two from our second factor. If we write them all up together, and we found that the solutions are gonna be equal to zero, X equals two pi divided by three X equals five pi divided by three and X is equal to two pi comparing this to our answer choices. A we can see that this corresponds with answer choice. A 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). (tan x - 1) (cos x + 1) = 0

Key Concepts

Trigonometric Functions

Zero Product Property

Interval Notation

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