Solve each equation in x over the interval [0, 2π) and each eq...

Question

Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.

3 tan 3x = √3

Accepted Answer

Welcome back. Everyone in this problem, we want to find the exact solutions of the equation three times two X equals a negative square root of three. And we're thinking about the interval, open bracket 02 pi closed parentheses for our answer choices. A says that it's 1/12 of pi 5/12 of pi 7/12 of pi and a 19 12th soft pi B says it's 5/12 of pi 11/12 of pi, 17 12th of pi and 23 12th of pi C says it's a third of pi four thirds of pi and seven thirds of pi. And D says it's 1/4 of pi, 3/4 of pi and seven fourths of pi. Now, if we're going to find the exact solution first, let's try to simplify our equation. 3, 10 X equals the square root of or the negative square root of three. Now that means if we divide both sides by three, the tangent of two X equals negative square root of three divided by three. So now we're asking ourselves what possible angle, the expression two X can we put into the tangent function to get a result of the negative square root of three divided by three. Well, since we are thinking about the exact solution, let's use the unit circuit to help us. Nor recall that on the unit circle where our unit circle basically just this circle that has, that goes from one to negative one on both axes. I recall that on the unit circle, the X and Y coordinates are written in the form cos theta sine theta. OK. Where cos theta is our X value and sine theta is our Y value. Now, although the coordinates aren't written in terms of the tangent, we can easily figure it out because we know that the tangent is actually the ratio of the sine function to the cosine function. So let's think to ourselves what Y value divided by an X value is going to give us a negative square root of three divided by three. Well, from our unit circle, we can recall that that angle, OK? That angle. Well, I'm good. Let me put that in right here would have been 5/6 of pi or one of those uncles would be 5/6 of pi. And we know that it's 5/6 of pi because at 5/6 of pi, the cosine value, the X value is negative square of three divided by two and the sine value is positive a half. And if we were to divide both of those, OK, then the tangent that we get. So we're talking about 5/6 of pie here, the tangent that we get would be equal to a half, right divided by a negative square root of three, divided by two, which is a half multiplied by a negative two divided by the square root of three. No two cancels two that leaves us with negative one divided by the square root of three. And if we rationalize by multiplying both by the numerate and the denominator by the square root of three, we would get the square root of three divided by three, which would be negative. So that's how we know the tangent of 5/6 of pi is equal to negative square root of three divided by three. There's another angle in the unit circuit that's equal to it. And that angle OK? That anger would be equal to 11 6th of pi. Let me draw my anger here. So that's 11 stick stuffed pie. And we also know that that is true because the point at that angle is equal to the square root of three divided by two and negative a half. Those are the coordinates. So for our unit circle, right, for our unit circle, we've seen that the tangent of 5/6 of pie, OK is equal to the tangent of 11 6 stuff pie, which is equal to the negative spirit of three divided by three. Well, what does that tell us? Well, that tells us then that our angle two X is going to be equal to that. So that means two eggs is going to be write it here. Two X is going to be equal to 5/6 of pie. And two X is also going to be equal to 11 6th of pi in the unit circle. But no, we wanted, we didn't just want it within the unit circle, but we wanted the general solution. OK. And for the general solution, since the tangent has a period of 180 degrees or in radiance, that's pi degree or, or in radiance that would have been pi radiance. Sorry. Then generally, OK. So we can make some space here and then say generally two X is going to be equal to 5/6 of pi plus pi N. OK? And, and two X is going to be equal to 11 6th of pi plus pi N. So now we need to figure out we need to solve what those possible values of XS are. OK? Well, no, for two X equals 5/6 of pi plus pi N. That means X write it below it. X is going to be equal to 5/12 of pi plus a half of pie in likewise. Well, before we go to the second one, let's stick with the first one. Let's try to find all the solutions. Now here N represents any integer. So we want to use and we want to find the value of X for all values of N within the interval from 0 to 2 pi. So let's start with N equals zero. Now, for N equals zero, then X is going to be equal to 5/12 of pi that we know for sure. Next for N equals one, OK? For, for, for N equals one, N equals one, then X is going to be equal to five thirds of pi plus a half of pi multiplied by one which is simply a half of pi. And that is going to be equal to 6/12 of pi. If we change a half of pi in terms of 12th, that's 6/12 of pi plus 5/12 of pi, which could give us 11/12 of pi. OK. Next for any calls, two, OK? For N equals two, we're gonna have eggs being equal to five thirds of pie. OK? 52, soft py plus a half of pi multiplied by two, which would simply equal pi and now 5/12 of pi plus pi is going to be equal to 7/12 of pi. OK. So we're going on, we're going on next. Is there a value for pi for, or is there a possible value for N equals three? Well, when we think about it, right? When we think about it here. OK, then for N equals three, we have X being equal to five thirds of pi plus a half of pi multiplied by three or three halves of pi. OK. Now three halves of pi we can think about that as 18 12th of pi and 18 plus five equals 23. So for N equals three, we'll get X to be equal to 23 12th of pi. So yes, there is a solution. But we can imagine then that for N equals four, if we go any higher, it's going to be beyond the two pi and thus will not be within our interval. So here, these are the solutions that we have so far. Next, let's try it for two, X equals 11 6th of pi plus pi N. If that's the case, that means X is going to be equal to 11/12 of pi. OK, plus a half of pi multiplied by N. Now, for N equals zero for N equals zero, we know that we're gonna get 11/12 of pi. OK, which we have already had that solution. OK? I'm gonna get here. Notice that we had that one for N equals one. We're going to get X being equal to 1112 soft pi plus plus a half of pi and 11/12 of pi plus a half of pi multiplied by one. That's gonna be 11/12 of pi plus 6/12 of pi, which would be equal to 17 12th of pi which again, we've already had that answer here for any cause too, for our first solution for N equals two, then X is going to be equal to 11/12 of pi. OK? Plus a half of pi multiplied by a pi. So that's 11/12 of pi plus pi and 11 plus 12 pi divided by pi. Eventually we're gonna get 23 twelves of pi. And I guess by now you see the pattern that here, that would be our original, a part of our original solution. And if we go beyond this, we know that it's going to be greater than two pi. So therefore, for this equation, our solutions are 5/12 of pi, 11/12 of pi, 17 12th of pi. Ok. And 23 12th of pi, if we look back in our answer choices, that would have been answer choice. B thanks a lot for watching everyone. I hope this video helped.

Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.

3 tan 3x = √3

Key Concepts

Solving Trigonometric Equations

Properties of the Tangent Function

Interval Notation and Angle Measurement

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