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). sin² θ - 1 = 0

Accepted Answer

Hey, everyone in this problem, we have the trigonometric equation seven minus 14 sine squared H is equal to zero. We're asked to use algebraic techniques to solve in the interval from 0 to 2 pi excluding two pi. And to give the answer in reg we have four answer choices. Option A and B are both sets containing three values for H. Option C is a set containing four values and option D is a set containing five values for H. We're gonna come back to those as we work through this problem. So let's start by writing the equation. We were given seven minus 14, nine squared H is equal to zero. Now, we don't want to get overwhelmed with the fact that we have a trike function this sign inside of our equation. We want to solve this just like any other equation we're trying to isolate H. So we want to get all of the numbers on one side and have this sine H on the other side. So we're gonna start by moving that 14 scored H to the right hand side, we give that seven is equal to multiplied by sine squared H dividing both sides by 14. And then we're just gonna flip the order that we're writing them in. So get that sine squared H is equal to seven divided by 14, which is just one half. And we can take the square root on both sides. And we get that sign of H is equal to plus or minus one divided by the square root of two. And this plus or minus is really, really important. And when you take the square root, remember that you'll get both of those roots. Yeah, once you write both of those you want to interpret based off your question and some questions, the positive root will make sense and some questions, the negative route will make sense. And in some both of them will, but you always want to consider them because it can be the difference between having just two solutions or maybe having four solutions. All right. So we have sign of H is equal to plus or minus one divided by the square root of two. What we want to do is think about our reference angle first. So the reference angle is gonna be an angle between zero and pi divided by two can be measured from the X axis. It satisfies our which and so if we think about drawing our special triangle, we wanna have the sides of a one in the squared of two. So we can relate them and we call it this special triangle. That has two shorter sides, each is one. And then the hypotenuse of the square root of two. OK. Because if you think about the Pythagorean theorem, one squared plus one squared gives you two, you take the square root to get the hypotenuse which is the square root of two. Now, we have a 90 degree angle in our triangle and we know that both of the shorter sides are the same length, which means that the other two angles must be the same. OK. So each of these must be pi divided by four, all right. So this is our special triangle. And if we think about sine being equal to one divided by the square to four, while the angle H must be pi divided by four, then OK. So we're gonna say that H prime, our reference angle is gonna be equal to pi divided by four. And when you're solving equations like this, you have to be careful not to stop there. OK. We have that sign can be positive or negative. So this solution could actually be in any of the four quadrants. So let's draw this out and see what we have. We're gonna draw out all four quadrants. Mhm Now in the first quadrant, we have this angle of pi divided by four or reference angle. OK. So we're measuring from the X axis in the counterclockwise direction pi divided by four in quadrant two. So the top left, OK. Remember our reference angle is always measured from the x axis. So in quadrant two, we're taking it from the negative x axis, we're going clockwise pi divided by four units divided by four radiant. Sorry until we reach the second potential solution quadrant three, the same thing, we're pi divided by four radiant from the nearest x axis. In this case, that's a negative X axis. But we're moving counterclockwise. And finally, in quadrant four, we're going from the positive X axis downwards in the clockwise direction high divided by fourth, right. So we have four possible solutions here and we need to figure out what the angle of each of them is. OK. We know the reference angle but what is the actual angle? And when we think about the actual angle, we want to think about taking it from the positive X axis in the counter clockwise direction. And that's how we note angles. All right. So each one is going to be the solution from the first quadrant and that's just gonna be our reference angle. So each one is gonna be pi divided by four. We're quadrant two. OK. We're gonna go from the positive X axis in the counterclockwise direction until we hit that solution. And this is gonna be our solution H two. Now age two. OK. If we imagine going all the way to 180 degrees, OK? Or pi radiance, that's gonna be the straight line, but we're a little bit short of that. And how short are we? Well, pi divided by four radiant and so each two is gonna be pi minus pi divided by four radiant. OK. We simplify this, we can write it as three pi divided by four. Now, when I mentioned that pi divided by four was going clockwise. OK. That kind of indicates because it's clockwise, it's a negative angle. That kind of indicates that we're gonna be subtracting. But I find the easiest thing here is just to draw it out and it makes it very clear which angles you should be subtracting which ones you should be adding in order to go from your reference angle to the actual angle. OK. So always draw the diagram if you can and then you'll be able to see more easily which ones to calculate. All right. So quadrant three, we do the same thing. We go from the positive X axis all the way around to this angle. Now we can see for H three, what's gonna happen is we're gonna go pi radiance. OK. So from positive X axis to the negative X axis, that straight line 180 degrees or pi radiance and then we're gonna go pi divided by four more. So H three is gonna be pi plus pi divided by four, which gives us five pi divided by four. And our final possible solution which is in quadrant four, we go all the way from the positive X axis around to quadrant four to that solution. We're gonna call this H four and we can see that H four and we go all the way around and complete the circle. That would be two pi radiant. But we're stopping pi divided by four radiant short. So H four is gonna be two pi minus pi divided by four radiant which gives us seven pi divided by four. So if we want to write our solution set, we found four answer choices. They are pi divided by four, three pi divided by four five pi divided by four and seven pi divided by four. If we compare this to our answer choices, OK? Based just off the number of solutions in our set, we have that C is the only possible answer. And if we compare the actual answers, we found two option C, we can see that that is in fact the correct answer choice. And so we have 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). sin² θ - 1 = 0

Key Concepts

Quadratic Form in Trigonometric Equations

Solving Basic Trigonometric Equations

Interval Restriction and Solution Sets

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