Exercises 39–52 involve trigonometric equations quadratic in f...

Question

Exercises 39–52 involve trigonometric equations quadratic in form. Solve each equation on the interval [0, 2𝝅). 9 tan² x - 3 = 0

Accepted Answer

Hey, everyone in this problem, we have a trigonometric equation 18 minus 6 10 squared M is equal to zero. And we're asked to use algebraic techniques to solve in the interval from 0 to pi excluding two pi. And to give the answer in radiance, we have four possible answer choices. Option A is that there is no solution. Option BC and D are all sets that contain four values for M and they just differ in which values those are, we're gonna get started by writing out what we were given 18 minus 6, 10 squared M is equal to zero. Now, don't get scared by having an equation with trade functions in it. We're gonna treat this just like another equation where we're trying to solve for M. Our goal is to get all the numbers on one side have M on the other side. We're gonna start by moving our 6 10 squared M to the right by adding, we have 18 is equal to 6 10 squared M dividing both sides by six. And we're just gonna reverse the side. So we can write it as P squared M is equal to three. And then we're gonna take the square root, OK? And we get that 10 M is equal to plus or minus the square root of three. And it's really important when you're working on trick problems to include this plus or minus. OK? Always remember when you take the square root, you get the positive and the negative root. So you want to include them and then infer from the problem whether one of them makes sense, the other makes sense or both makes sense. OK. In this case, we're gonna have both of them. Now, we're looking for values of M or tangent of M is equal to plus or minus the square root of three. If we were solving this to get a decimal value, we could take the inverse can on both sides that would give us our value. But we want to solve this exactly in radiance. So let's think about our special triangle and we're gonna think about finding a reference angle first. OK? So we're gonna think about T and M is equal to a square root of three. Now, let's draw on our special triangle and it's a really good idea to try to remember the special triangles if you can. Now the one we need when we need to have a side length of the square root of three. And the special triangle that has a side length of the square root of three has side lengths, one in the square root of three with the hypotenuse of two. And you can always double check by using Pythia in theorem, we have one squared plus the square root of three squared gives us four. We take the square root, we get two which is a hypo. And now this is a right angle triangle. The other two angles must set up to 90 degrees or pi over two across from the smallest side which is one, we're gonna have the smallest angle. OK. And you want to recall that this angle is gonna be pi divided by six for this special triangle. And the other angle is gonna be pi divided by three. So we're talking about tangent. The tangent we want to recall is gonna be the opposite side versus the adjacent side. OK. So the angle that gives us that the opposite side divided by the adjacent side is the square root of three is gonna be pi divided by three. So we have M prime, our reference angle is gonna be equal to pi divided by three. Now, we have tangent can be positive or negative, which means it can be in any quadrant that we have. So we're gonna go ahead and draw all of our quadrants. We're gonna draw the reference angle in each quadrant and then we're gonna work out what the actual angle is for each of those quadrants. So let's go ahead and do that. We have our Cartesian plane here and recall that our reference angle is gonna be the angle between the x axis and the angle in that quad. OK. So in quadrant one, we have the angle between the positive x axis and this angle is gonna be pi divided by three. The next one we have at that angle and quadrant two is gonna go from the negative x axis. OK. Where it's always just going to the from the x axis in whichever quadrant we're in. And so in this case, we're going from the negative X axis upwards pi divided by three radiant quadrant three A again, the negative X axis. And here we're working down pi divided by three radiant. And finally, in quadrant or we're going from the positive X axis downwards. OK? We have these four possible angles. But remember when we're giving solutions to our problem, the angles are always measured from the positive X axis and then counter clockwise. OK. So our first angle which we'll call m one is gonna come from quadrant run. And that's just gonna be our reference angle pi divided by three because we can see that we go from the positive X axis up pi divided by three radiant. That's one of our solutions. The next one, we're gonna go from our positive X axis, we're gonna counter clockwise until we hit the angle we have in quadrant two. OK. So the solution we have in quadrant two, we're gonna draw in green and we're gonna call it X two, oops not X two M two M two. Now, you can imagine that if we continue this green down to the negative X axis and we'd be going along this straight line from the positive to negative X axis. That would be pi degrees or pi radiant sorry or 100 and 80 degrees. But we're stopping short, we're stopping pi divided by three radiant short. OK. So what M two is gonna be equal to is it's gonna be equal to pi and that total amount going all the way to the negative X axis minus the reference angle of pi divided by three. And this gives us an angle of two pi divided by three. OK. So that's quadrant two. We're gonna do it again. For quadrant three, we start on the positive X axis. We go counterclockwise all the way around until our solution in quadrant three. We're drawing this in blue this time and we're gonna call this M three. What we can see now is that we've gone all the way around to the negative X axis. So that's a pi radiance. And now we're going our reference angle pi divided by three more. So M three is gonna be equal to pi plus pi divided by three. And this is why I like to draw odor Cartesian plane, draw out these angles. I think it makes it a lot easier to tell whether we should be adding our reference angle subtracting our reference angle, whether we're doing it from pi or from two pi. And I think the drawing really helps. So if you ever get confused on whether you should be adding or subtracting, draw out your angles and see what you find. So M three, we have pi plus pi divided by three, which gives us four pi divided by three. And finally, we have our solution in quadrant four, we're gonna go from the positive X axis all the way around counterclockwise until we hit that solution. We're gonna call this M four. Yeah. Now M four, if we go all the way around back to where we started, that's gonna be two pi and that's an entire circle. But we stopped pi divided by three short of that. OK. So our angle M four is gonna be equal to two pi minus pi divided by three. And we can write two pi as six pi divided by three. We subtract pi divided by three and we get five pi divided by three. And so we found that in this case, there actually are four solutions. OK? So we're looking at answer choice, either BC or D. We found that these solutions are pi divided by 32 pi divided by three, four pi divided by three and five pi divided by three, which corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

Exercises 39–52 involve trigonometric equations quadratic in form. Solve each equation on the interval [0, 2𝝅). 9 tan² x - 3 = 0

Key Concepts

Quadratic Form in Trigonometric Equations

Solving Quadratic Equations

Finding Solutions on the Interval [0, 2π)

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