Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College solving trigonometric equations. When we have a trigonometric equation we want to get the trig function by itself. So in this case, if I subtract 1 from each side and then I divide by a negative, I want to know where does cosine Theta equal 1/2? Well, cosine Theta equals 1/2 at two different locations. If we think about our cosine graph, cosine is going to be 1/2 at two different locations. It's going to be here when it's decreasing on the function and again here when it's increasing on the function. So Theta would equal π thirds when it's coming down on the function, but it repeats over here when it's coming down on the function again. So to hit every possibility, we're going to say π thirds plus 2K Pi where K is some integer. Well, where else is cosine Theta 1/2? Well, it's also 1/2 here when the function is increasing, so that's up five Pi thirds. Well, it's also true here where it's increasing and back here where it's increasing. So we're going to add 2K Pi to get all possible values. Now, if the direction said we only want values that are less than 2π and greater than or equal to 0, what we would then do is we'd actually say, well, what if K = 0? If K = 0, we'd have π thirds and we'd have 5 Pi thirds. And then we'd think, well, what if K equal 1? Well, in this case, if K equal 1, we'd have 7 Pi thirds, and that would be too big. That'd be bigger than 2π. We'd also have 11 Pi thirds, and that one's bigger. So in a cosine graph, cosine Theta equal 1/2. In a regular. That occurs twice. So our answer would be π thirds and five Pi thirds. Looking at the next example, we're going to add the 1 and divide by 2 so we get sine squared Theta equal 1/2. Well, when we square root something we remember we have positive and negative and positive negative sqrt 1 half is really positive negative root 2 / 2. So we want all the locations when sine is positive and negative root 2 / 2. If we think about our sine graph, our root 2 / 2 is going to occur twice when it's positive and twice when it's negative on a regular zero to 2π. So Theta is going to be π force plus 2K Pi for a positive, and theta's going to be 3 Pi force plus 2K Pi for the positive root 2 / 2. For the negative root 2 / 2, we'd have Theta equaling 5 Pi force plus 2K Pi, and we'd have Theta equaling 7 Pi force plus 2K Pi. Now if we wanted only the values in between 0 less than or equal to Theta less than 2π, we'd put in K = 0 and we'd get π fours, 3 Pi fours, 5 Pi fours, and seven Pi fours. If we put in K equal 1 because we're adding 2K Pi each time, we'd be too big. So our 4 answers here would be π fours, 3 Pi fours, 7 Pi fours, and five Pi fours, 3 Pi fours, 3 Pi fours, 5 Pi fours, 7 Pi fours. This next example, when is sine -1? Well, sine is -1. And this problem, it's going to be when 3 Theta equal 3 Pi halves, but it occurs every cycle or every period. So plus 2K Pi. In order to get the Theta by itself, we're going to divide by three and we're going to get Theta equaling Pi halves plus 2/3 K Pi. Well, this time, if we were, that's going to be every possible value. If we're wanting them all the values that are within one between 0 and 2π, we'd let K = 0 and we get π halves. If we let K equal 1, we would get π halves plus 2/3 Pi. So if we have π halves plus 2/3 Pi, we'd have to get a common denominator. So we'd get 3 Pi 6 + 4 Pi 6, and that's less than 2π. If we let K equal 2. Next we get π halves plus 4 thirds π getting a common denominator again, three Pi 6 + 8 Pi 6. That's 11 Pi 6, and that's also less than 2π. What if K equal 3? Well, we'd have Pi halves plus 2π. That one's going to be too big. So we're going to have three values Pi halves 7, Pi 6:00 and 11:00 Pi 6. Now one of the things that we can think about is this 3 on a regular. Sine equals -1 one time. The three times the Theta tells me we should expect it three times. Let's look at some more examples. Tangent of two Theta equal -1. Well, when does tangent equal -1? When the angle is 3 Pi fours, but tangent. Occurs every K π. So if we wanted Theta, we'd have three Pi eights plus K Pi halves. That would give me every single possibility. If I wanted only the ones that are in between 0 and 2π, I'd put K = 0 and I'd have three Pi eights K equal 1 and I would get 3 Pi eights plus Pi halves. Getting a common denominator that Pi halves is 4 Pi eights, so we'd get 7 Pi eights, K equal 2. We'd get 3 Pi eights plus π, which would be 11 Pi eights. If K equal 3/3 Pi eights plus three Pi halves, getting common denominator again, so that three Pi halves is going to turn into 12π eights, so 15 Pi eights. That's less than 2π. If we tried K equal 4, that's going to be too much. Three Pi eights plus 2π. So my 4 answers are going to be 3 Pi eights, 7 Pi eights, 11 Pi eights, and 15 Pi eights. If we were looking specifically in between 0 and 2π. If we were wanting all possible answers, our answer would have been three Pi eights plus K Pi halves. If you thought about how many times is tangent -1 in a unit circle, it's twice. But that two times the Theta would tell me to expect 4 answers. Well, how about tangent Theta over 2 equal negative root 3, so Theta over 2? When is tangent root 3 tangents root 3 at π thirds plus K π because it occurs every period. So Theta is 2π thirds plus K π. That's all possible answers. Well if we wanted only in between 0 and 2π, we'd say let K = 0 and we get 2π thirds. If we let K equal 1, we'd get 5 Pi thirds. Oh no we wouldn't come back here a second. When I multiply, I need to multiply by two, so that's 2K Pi. So now if K = 0, we get 2π thirds. But if K equal 1 were too much, we'd have 2π thirds plus the 2π. Now if we think about this, how often is tangent root 3 in a unit circle? And that occurs twice. But now our angle is divided by two, so we should expect only one answer, which is what we got. Cosine 2, Theta equal negative 1/2 when is cosine negative 1/2. Well, cosine is negative 1/2 when the angle is 2π thirds plus 2K Pi, but also when we're at 4 Pi, thirds plus 2K Pi. If we divide by two, we'd get Theta equaling π thirds plus K π and we'd get Theta equaling 2π thirds plus K Pi. Those are all possible solutions. If we wanted only in between 0 and 2π, we would then say let K = 0. If K = 0, we'd have π thirds and we'd also have 2π thirds. If K equaled one, we'd have 4 Pi thirds and five Pi thirds. If K equaled 2, this would be 2π and it would be past our zero to 2π restriction. Another way to think about it is -1 half. How often does cosine have negative 1/2 in a unit circle? It does twice, so that two times the twice should expect 4 answers. This next one. When is sine one? Well, sine is one at π halves, so 3 Theta plus π eighteenths is going to equal π halves and it occurs every two K π. Now we want to solve for Theta, so 3 Theta is going to equal π halves minus π eighteenths plus 2K Pi. We're going to be able to only combine like terms, so the one with the K we can't combine, but the Pi halves minus the π eighteenths we can combine Pi halves is really 9 Pi eighteenths. If we take 9 Pi eighteenths and -1 Pi 18th, we get 8 π eighteenths and eight Pi eighteenths is going to reduce to four Pi ninths. Now, to get the Theta by itself, we're going to divide by 3. So we get 4 Pi 20 sevenths plus 2/3 K Pi if K = 0. That's all possibilities. Now if we want only in between 0 and 2π, we're going to have K = 0, so we'd get 4 Pi 20 sevenths. If K equaled one, we have to get a common denominator, so 27. So 2/3 would turn into 1820 sevenths. 1820 sevenths plus 420 sevenths would make 22 20 sevenths Pi. If K equal 2. We'd have 4 thirds π + 4 Pi, 20 sevenths 4 Pi thirds +4 Pi. 20 sevenths would give me 40 Pi 20 sevenths if K equal 3. This ending portion, the 2/3 * K Pi would be 2π and that would be too much in a unit circle. How often is sine one? It only occurs once, so that three would tell me I should be expecting three times that many answers, or four Pi 20 sevenths. 2220 sevenths Pi 40 Pi 20 sevenths. The next example, we have two terms that are going to multiply to give me 0 so that we know that each of those terms individually would possibly equal 0. So if we look at cotangent Theta plus one equaling 0, when is cotangent Theta equaling -1? And that's occurring at three Pi force plus K π because cotangent. Is π. When is cosecant Theta equaling 1/2? That's not possible because cosecant has to be in between negative Infinity and -1 or one to Infinity. So if we want all the possibilities between 0 and 2π, we'd let K = 0 and get 3 Pi force. We'd let K equal 1 and we would get seven Pi fours. If we have K equal 2, we're past 2π. Looking at a couple more, this one we're going to factor first. So we're going to get 2 cosine Theta and cosine Theta, and we're going to have a -4 and a +1. If we foiled that out, we'd get negative. We'd get the original equation. So cosine Theta is equaling negative 1/2 from the first parentheses. Cosine Theta is equaling 4 from the second. Cosine Theta is never going to equal 4 because it has to be in between -1 and one. Cosine Theta is negative 1/2 when Theta equals 2π thirds plus 2K Pi, or when Theta equals 4 Pi thirds plus 2K Pi. Those are all possible answers. If we only want the answers between 0 and 2π, we start by letting K = 0 and we get 2π thirds and we get 4 Pi thirds. If K equal 1, we would be too far because we'd have a 2π at the end being added each time. This next one we have to figure out how to get everything in terms of one trig function. So instead of cosine squared Theta, we're going to use our Pythagorean identity that said cosine squared Theta plus sine squared Theta equal 1. So if cosine squared Theta plus sine squared Theta equal 1, we know that cosine squared Theta is 1 minus sine squared Theta. Then we have a minus sine squared Theta from before and a plus sine Theta all equaling 0. If we distribute, we get -2 sine squared thetas plus sine Theta -1 + 1. That's just combining our like terms. I don't personally like the leading coefficient to be a negative, so I'm going to multiply the entire equation through by a -1 to make that first term positive. So we're going to get 2 sine Theta and sine Theta and a -, 1 here and a + 1 down here. So sine Theta is going to equal 1/2 and sine Theta is going to equal 1. Well, when a sine Theta 1/2, sine Theta is 1/2, when Theta equals Pi 6 + 2 K π. Also when Theta is five Pi 6 + 2 K π. When is sine Theta equaling 1, sine Theta equals one for Theta equaling Pi halves plus 2K Pi. So that would be all possible answers if we only wanted the answers in between 0 and 2π. We're going to start by letting K = 0, and we'd get PI65, Pi, 6 and π halves if K equal one. That 2π at the end would make them all bigger than 2π. The last example, we have to remember that sine is an odd function. So sine of negative Theta is really going to be the same thing as negative of sine Theta. That's due to the fact that sine of negative Theta is or sine Theta is an odd function. So sine of negative Theta is really the same thing as negative sine Theta. When we look at this, those negatives are going to cancel and we're going to get cosine equaling sine. If we thought about dividing each side by cosine, we'd get one equaling tangent because sine divided by cosine is tangent. Well, when is tangent, one tangents one at π force plus K π. So if we have π force plus K π, that's all possible answers. If we only wanted the answers in between zero and two Pi K = 0, we'd get π force. K equal 1, we'd get 5 Pi force K equal 2 and we would be too big because of the 2π at the end. Thank you and have a wonderful day.