 We're now going to talk about a few specific examples of trigger equations. These are special cases. The first is a trigger equation that is quadratic, quadratic in form. This one can actually be solved using the square root method. So isolate the quantity squared, which would be cosine. Add one to both sides. Divide by two on both sides, and then you'll take the square root of both sides, and you will get cosine of theta equals plus or minus square root of one, which is one, square root of two, which is square root of square root of two. So we have cosine positive and negative, which means, I'll write it down here, since cosine is positive and negative, we want theta in all four quadrants. So it's quadrants one, two, three, and four. How exciting. So first, let's find out what that reference angle is going to be. What is our mystery reference angle going to be? A question mark. Organa, if you like to use the Greek notation. All right, so we have cosine is one over square root of two. Adjacent as one, hypotenuse is square root of two, which means my other side is one. A 45, 45, 90 triangle. My reference angle here is going to be pi over four. We have to move this angle to all four quadrants. So first off, quadrant one. Well, good news. It's already there, right? Quadrant one solution is data is equal to pi over four, quadrant two. Given a reference angle, how do I move it back to quadrant two? Well, my angle theta is going to be pi minus my reference angle. That'd be three pi over four. All right, quadrant three time. This one's my favorite actually, because we just take the reference angle and we add pi to it. Gives us five pi over four. So I have one pi over four, three pi over four, five pi over four, and what would likely come next? Well, if you haven't figured it out yet, theta is equal to two pi. Minus pi over four, which is indeed seven pi over four. Four solutions here. Pi over four, three pi over four, five pi over four, and seven pi over four. These are the four solutions in the interval zero to two pi. Keep in mind the original trig function here had no adjustments to its period, so the period is just two pi. So adding two pi to any of these values would clearly get you outside of the interval zero to two pi. Next example, which is still quadratic in form requires a little bit of factoring. So just to get this in a more familiar format, everywhere I see sine theta, I will replace it with x. So I'll have two x squared minus five x plus three equals zero. We may like this because it brings us back to those basic algebra days. So factor two x squared minus five x plus three. You will get two x minus three and then x minus one equals zero. Take each piece, set it equal to zero and solve for x. So first piece when you solve for x, you get x equals three halves. And you get x equals one. Remember what x is? x is sine theta. So sine theta equals three halves or sine theta equals one. So where is sine equal to three halves? My basic sine function, where is it equal to three halves? And the answer is actually nowhere because sine theta only spits out values between negative one and one. Sine of theta must be between or in the interval negative one to one. So sine of three halves is not true. It is undefined. So sad we can't do anything with this first piece. Second piece, where is sine theta equal to one? Well, you could write this as one over one, but realize you can't have an opposite side and a hypotenuse, a leg and a hypotenuse have the same length. That would be one messed up triangle. So draw your unit, circle and try to find out where the second coordinate of the ordered pair is one. And that would be up top here at pi over two. Cosine of zero, sine is one. Sine is equal to one at pi over two. So theta is pi over two. So all that hard work and we actually only obtained one solution, theta equals pi over two. But we did have a little bit of factoring fun in the process of getting there. Next example is not quadratic in form, but requires the use of some trig properties that you've learned previously. So these properties will never go away. I can promise you that. And the first thing you want to do is negative sines and side sine functions. They are able to be brought outside. Why? Because sine is an odd function. Next, your goal is to get all of your theta terms to one side. So we can divide both sides by sine theta. And divide both sides by sine theta. And what is cosine over sine? Cosine over sine. That'd be cotangent. So I get cotangent theta on the left side and negative one on the right hand side. So I have cotangent theta is equal to negative one over one because I do like to have a ratio here. But keep in mind that since cotangent is negative, theta is in what quadrant? Quadrants two and four. Because cotangent and tangent are positive in one and three. So cotangents negative and quadrants two and four. Let's draw a triangle. We have our mystery angle in the bottom left. Cotangent is one over one, one one, square root of two. And it looks like my angle is going to be pi over four. My mystery angle, my reference angle is pi over four. And I have to move this to quadrant two and to positive quadrant four. So how are we going to do this? We're going to use some formulas. That's what we're going to do. So first off, we'll start with quadrant two. The angle theta would be pi minus the reference angle, three pi over four. And then we'll move to quadrant four. My solution for theta, my quadrant four angle would be two pi minus pi over four, which is actually seven pi over four. The cool thing about 45 degree angles is quadrant one's pi over four. Two is three pi over four. Quadrant three is five pi over four. And quadrant four would be seven pi over four. It's kind of cool how this works out. So the solutions for theta would be three pi over four, seven pi over four. Once again, the period is two pi of our trig function. So adding two pi is not going to keep us in the interval zero to two pi. It will take us out. So there's only two solutions in the interval zero to two pi. This last one, let's use a calculator. Now what you have to do for your calculator input here is you have to use inverse sine functions. So you'll have to input inverse sine of 0.4. Make sure you're in degree mode or radian mode, whatever it needs to be in. Typically you want to be in radian mode for this section. So if I plug in sine inverse of 0.4 and I'm in radian mode, it's going to split, spit out a radian measure of 0.41. It's going to split out a radian measure of 0.41. Remember inverse sine is only defined for quadrant one and negative quadrant four. So since in this case sine was positive, it spit out the quadrant one angle. But sine is equal to positive 0.4 in many quadrants. Since sine is positive, theta would be in quadrant one and two. But the calculator is not going to give you your quadrant two angle because as far as it knows sine inverse is only defined at quadrants one and negative quadrants four. It doesn't realize that originally you just had a trig equation that was without any inverse trig functions. So I will want to move to quadrant two, move this angle to quadrant two. I do pi minus 0.41. You can use the pi key in your calculator or you could use, in my case I just used 3.14. So 3.14 minus 0.41 is going to give you 2.73. So it's theta in this case is the easy solution to get would be 0.41. And then the one that requires a little bit of thinking involved would be the 2.73. A lot of people always forget about the 2.73. Remember the inverse trig function was only used as an intermediate step. It was not part of the original question. So I can open up all four quadrants. In this case quadrants one and two were the positive quadrants for sine. Well that's all I have for you right now. So I hope you enjoyed. Thank you for watching.