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). 4 cos² x - 1 = 0

Accepted Answer

Hey, everyone in this problem, we have the trigonometric equation nine minus 18 cosine squared of G is equal to zero. And 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 radiance, we have four answer choices. Option A B and C for all sets containing four values for G and they just have different values for each of those options. And option D is that there is no solution. So let's start by writing what we were given nine minus cosine squared of G is equal to zero. Don't get confused when you have an equation like this with trick functions. We want to solve it exactly the same way that we do with regular equations that we're used to. OK. We want to get all of the numbers to one side and that value G that we're looking for on the other side. OK. So we're gonna start by moving our negative 18 cosine squared G to the right hand side by adding it, we have nine, it is equal to 18 cosine squared G and we're gonna divide both sides by 18. And I'm just gonna reverse what side we're writing each of these on. So we have the cosine squared O G gonna be equal to nine divided by 18, which is one half. Then we take the square root and we get that cosine of G is equal to plus or minus one divided by the square root of two. OK? And the plus or minus is really, really important. And when you're working with square roots, you always want to remember that when you take the square root, you get both the positive and the negative route. And particularly when you're working with tri equations, this can be the difference between having no solutions, having two solutions, having four solutions. You can really change the solutions you're getting. So always remember that you have both roots and then you can interpret which one makes sense for your problem. Maybe you're only taking the positive route or the negative route. You always want to think about that when you take the square root. OK. So we have cosine that's gonna be plus or minus one divided by the square root of two. Now, the first thing we wanna do is figure out our reference angle. OK. So we want to figure out the angle between zero and pi divided by two that satisfies this equation. Hm So let's think about our special triangles, right? And the special triangles are really, really helpful. It's a really good thing to remember, so we have a right angle triangle. And we can see that our cosine is one divided by the square of two. So we want a special triangle that has side lengths of one in the square root of two. And it turns out we have one with both of those. So that's perfect. OK. If we have a special triangle where the two smaller sides are both equal to one, then the hypotenuse is equal to the square root of two. And if you think about Pythagorean theorem, one squared plus one squared is equal to two, take the square root to get the hypotenuse, we get the square root of two. Now, for our angles, we already know that we have one angle of 90 degrees. OK? Or pi divided by two radiant. That means the other two must also add up to pi divided by two because it's a triangle, they must all add up to pi our two smaller side links are the same. So the angles must be the same. And so both of these angles are gonna be pi divided by four radiant. All right. Yeah. If we think of cosine, we have that cosine as one divided by the square root of two. OK. Well, cosine is the adjacent side divided by the hypotenuse. And so the angle that corresponds to that is pi divided by four. So our reference angle which we're gonna call G prime, it's gonna be equal to pi divided by four. And this is our angle less than pi divided by two that satisfies this equation. And we now we need to think about where we have cosine either positive or negative, where can we have solutions? And in this case, we have that cosine is positive or negative from taking that square root. So this can be in any quad, any cosine positive or negative means any quadrant we want. So let's go draw a diagram throwing our reference angle in every single quadrant and see all of the potential solutions. We have our playing here and we have our reference angle pi divided by four. And that's always measured from the X axis closest in the quadrant. So in quadrant one, we have pi divided by four coming from the positive X axis. In quadrant two, we have pi divided by four coming from the negative X axis in quadrant three. It's also gonna be coming from the negative X axis. And finally, in a quadrant four, we have pi divided by four coming from the positive X axis downwards. Now, what we see is that we've done four solutions here, OK. All of these have cosine equal to plus or minus one divided by the square of two. Now we need to figure out what angle this actually represents. OK, not the reference angle but the actual angle. And remember when we're talking about angles, we want to measure from the positive X axis counting counterclockwise. So we're talking about the first one in quadrant one, G one. Well, that's just gonna be equal to our reference angle. OK. In quadrant one, the angle is equal to the reference angle, we get that G one is equal to pi divided by four. When we're thinking about quadrant two, we need to go from the positive X axis counterclockwise to that solution. We're gonna call this G two. And what you can see is that if we go all the way to the negative X axis, that's a straight line that's gonna be pi radiance. But we actually stopped pi divided by four short of that. OK. And so we're angle G two, it's gonna be equal to pi radiance minus pi divided by four gradients which is equal to three pi divided by four. OK. So we're done quadrant one, quadrant two moving to quadrant three. And we're doing the same thing from the positive X axis all the way around counterclockwise to our solution. We're gonna call this G three. We've shown it in green. Now, we can see we do make it to the negative X axis pi radiance and then we go that reference angle further. So we go to pi radiance and then we go pi divided by four radiant more. And so G three is gonna be equal to pi plus pi divided by four. OK. And this is gonna give us five pi divided by four. And finally, we have Q four quadrant four. We're going from the positive X axis all the way around counterclockwise until we hit our solution. We're calling it G four. We can see that if we go all the way around to back to where we started, that would be two pi radiant, but we're stopping pi divided by four radiant short of that. And so our angle G four is gonna be equal to two pi minus pi divided by four radiant which gives us seven pi divided by four. And now we've written out the solution or all four of our solutionss using this diagram. And I find these diagrams really, really helpful and sometimes it can be confusing whether you should be adding or subtracting your reference angle, whether you should be using pi or two pi. So I find if you draw it, draw on it, it really helps you visualize and know exactly what you should be adding subtracting and from which angles. All right. So we have our four solutions here. We found that pi divided by four, three pi divided by 45 pi divided by four and seven pi divided by four are all solutions to this equation. If we go up to our answer choices and compare what we found, we see that this 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). 4 cos² x - 1 = 0

Key Concepts

Quadratic Form in Trigonometric Equations

Solving Basic Trigonometric Equations

Interval Restriction and Solution Sets

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