I add this question to every trig test

this is a great test question the reason why this is a great test question is because it's going to challenge you on factoring and one thing I know about trick is once we get into trick someone's like fine away from like algebra and factoring and all the other stuff and then guess what bam we come back with some more algebra so when we're trying to solve a trigonometric equation typically we're not going to be doing factoring and stuff like that but in this problem that's exactly what we're going to need to do a lot of times we can simplify a trigonometric equation by using some identities or inverse operations to simplify it so therefore we can go and solve but in this example notice that we have a cosine squared of X and A cosine of x that's something you want to look out for and when you see that what I want you to do is use substitution okay so sometimes we use you in this case I'm just going to use x because X is the most famous variable that we've used in our algebra so if I'm going to say cosine of x is going to be equal to X now what I'm going to do is I'm just going to rewrite this equation with x's so remember that's a cosine Square that's the same thing as cosine of x quantity squared right so that's why it's an x square there and now you can see we have a basic algebra one problem guys right even though we're doing trig it's an algebra one right I mean this is just a typical solve the quadratic and the first thing you need to know when solving a quadratic right when we have an x squared and an X is get everything equal to zero it amazes me as a teacher how many times will take like a week or two break and students have already forgot everything they've learned about algebra polynomials quadratics and stuff like that right that's the number one step now I know a lot of you want to sometimes default to quadratic formula and stuff like that once you learn it it's like why do anything else right or type it into chat GPT and just go ahead and solve it but this really isn't that bad of a problem whenever you have a coefficient you know especially of like two three or four a lot of times guys you can do these rather quickly faster than it'd be for you to take out your calculator or your phone and try to figure out the answer well if you know what we want to factor this and it's going to give a binomial squared and we're going to have a 2 in front here you can do this in your head 2X and X we know my last two numbers have to add to multiply to give us one so it's either plus one and plus one or negative 1 and negative one since my middle term is positive guys look at this there's only one answer like that was the easiest Factor problem ever but you'd be amazed by how many students that are really really bright know how to do all this trick will get stumped on a problem like this now again in this case we can apply zero product property and we could solve but at this point I think we did the hard work with the factoring right so now we're going to do is I'm going to go back to my substitution and I'm just going to go ahead and rewrite that you know what let's go to 2X plus one equals zero X plus one is zero okay so now we apply the zero product property now let's go back and rewrite this in this factored form from the zero product property I lied I don't know why as a teacher I'm like I want to write this in factored form I think could anybody that's like that would go back through this and look at it I want you to go from non-factored form to factored form right so we add the one to the other side and make sure you do that and now let's go and read three write these in zero product property okay so now we can just use our inverse operation to solve right what would you do over here subtract one divided by two what would you do over here subtract one divide by two what about over here what would you do subtract one on both sides what are you over here subtract one on both sides you see you see how the importance of everything we did in algebra class is now preparing you for applying this with trigonometry now the one thing you didn't learn in your algebra class was to solve these equations using the unit circle now remember when we're trying to we're saying the cosine of x we're saying the cosine of an angle which represents the x coordinate of the coordinate point on the unit circle and thankfully I recognize negative one-half and I recognize negative one now a lot of times we try to find the values on a restriction from like zero to two pi okay and that basically is going to be just the unit circle so let's go and figure out what are all the solutions where I'm going to have cosine of x is equal to a negative one-half now hopefully you know you're in a circle and you know those two angles are going to be a 5 pi over 6 and a 7 pi over 6. then we look at negative 1. and that's going to be up here and that's going to be your answer negative pi okay now if we needed to find all the solutions right what I want you to recognize is here's one solution that is the angle five pi over 6. okay the next angle is you're gonna have to add 2 pi to it and you just keep on adding to Pi so for all of these answers if I wanted to find all the solutions all you're going to do is just say plus 2 pi n plus 2 pi n plus 2 pi n you can do that for any single angle any single time you're trying to convert them to all solutions the one caveat though is a lot of times when they have a when they're only like Pi away or something like that you can sometimes simplify that so what you always want to look for a more simplified solution just add your 2 by n so hopefully this video was helpful for you guys and if it was I know you're gonna join the next video