 Welcome back to our lecture series math 1060 trigonometry for students at Southern Utah University. As usual, I'll be your professor today, Dr. Anjan Misildine. In lecture 21, we started to talk about solving trigonometric equations. We focused on linear and quadratic trigonometric equations. In lecture 22, we're gonna continue to develop these methods. Now, in addition to the solely algebraic methods we introduced in lecture 21, in lecture 22, we're gonna also use trigonometry to help us solve trigonometric equations. In particular, we're gonna use trigonometric identities to transform the equation into a linear or quadratic equation involving sines or cosines that we saw previously in lecture 21. Basically, we're gonna use identities to convert the equations we see right now into equations of the format we saw in the previous lecture, so that we're in a better situation to solve them. So for example, let's say we wanna solve the equation to cosine x minus one equals sine, a secant of x, excuse me. We need to do this on the interval zero to two pi. So we wanna find all solutions from zero to two pi. Clearly, we're looking for solutions which are in radians. That's an important thing to remember there. We don't need the general solution just up to two pi there. How are we gonna solve this equation? Well, we have a cosine, right? We also have a secant. How are we gonna deal with this? Well, it's important to recognize some of our fundamental identities that secant is the reciprocal of cosine. Secant is one over cosine. So if I make that substitution, I can convert the secant into a cosine. And so we get the equation two cosine x minus one equals one over cosine of x, like so. For which now we have fractions in play here, but whenever you have an equation, fractions don't have to be your nemesis. You can always clear the denominators. So we don't want the fraction on the right-hand side, so we're gonna times the right-hand side by cosine of x, the denominator. But whatever is good for the goose is good for the gander. That is to say, whatever we do to one side of the equation, we have to do the same thing to the other side of the equation so that equality is preserved. So we times the right-hand side by cosine of x to clear up the denominator, we have to then multiply the left-hand side by cosine of x then distribute it through. And we end up with two cosine squared x minus cosine of x, this is equal to one. If we subtract one from both sides, we'll just move that over there. We end up with two cosine squared x minus cosine x minus one equals zero, like so. And so we're all to turn this into a quadratic equation in terms of cosine of x. For which then we could try to solve this thing by factoring. We could try to solve it by the quadratic formula. And when it comes to factor, there's a lot of techniques one could use here, the reverse foil method we've seen before. This one I think I can probably just kind of guess my way through it. Cause after all, how do you get a cosine squared? Well, you have to have a cosine times a cosine, right? How do you get a two cosine squared? Well, one of them has to be two. The other one has to have a coefficient of one. How are you gonna multiply together and get a negative one? Well, you have to have one times one, in which case it's gonna be positive, one's positive, one's negative. And then this has to combine together to be negative, negative cosine. So we need to have like a negative two. So let's do something like this negative positive, kind of guess your way through that. Again, you can use the reverse foil method as you've seen before. And just if you're gonna guess, you have to check it of course, two cosine times cosine is two cosine squared. You get negative one times positive one, that's a negative one. Then you're gonna get a two cosine plus a cosine. So that's negative two plus one, which is a negative one. There we have it. This is the correct factorization. Once you factored, of course, we then have to treat the two separate pieces. So there's the first possibility that two cosine X plus one equals zero. So if we think about that, two cosine X plus one equals zero, we get two cosine X is equal to negative one, divide that by two cosine of X equals negative one half. If you're negative cosine, that means you're in the second quadrant and the third quadrant. Cosine of X equals one half, exactly, again, just think of the first quadrant. That happens at a six degree angle, but we're working in radiance. We're gonna say pi thirds. This is a reference solution, right? So I just kind of did that for scratch work if you're thinking about this. In which case, then we get X equals, well, what is cosine equal to negative one half? That's gonna happen in the second quadrant when you reference pi thirds, which would be two pi thirds. And then the third quadrant when you reference pi thirds, which is four pi thirds. So that's that possibility. The other possibility is if cosine of X minus one equals zero. So if cosine of X minus one equals zero, that means cosine of X is equal to one. And that, of course, happens for cosine at X equals zero itself. Again, we're just going from zero up to two pi, but not included two pi. We don't need the general solution. And so these are the three solutions we end up with here. We get X equals zero, we get X equals two pi thirds, and we get four pi thirds as the three numbers that solve this quadratic equation, like so.