 In this video, we want to solve quadratic trigonometric equations where the period has been changed. Notice in this example, we have 2 sine squared of 3 theta minus sine 3 theta minus 1 of 0, and we want to solve this for all possible degree solutions, okay? Now, fortunately for us, the angles in play here are the same. We have 3 theta, 3 theta, because there's no disconnect with the angles like here's a theta, here's a 2 theta, here's a theta half. So we don't have to worry about those identities there. And also, since we have a sine squared and a sine, there's no mismatch of trigonometric functions. So again, no trigonometric identities are really going to be needed here. This is just a quadratic trigonometric equation for which the period has been changed. The period change we won't worry about until the end of the problem. So let's just solve the quadratic equation right now. We can do that by factoring, completing the square quadratic formula. Looking at the coefficients here, we have a 2 and a negative 1. Those come together to take 2 times negative 1, which is negative 2. I need factors in negative 2 that add to negative 1. That is this negative 1 right here. And we see that if you take negative 2 plus 1, that gives you a magic pair. So I could factor this thing. I could use the reverse foil method. Although, since the leading coefficient is 2 in the constant is negative 1, there's really not a lot of ways to factor 1 and 2. So I bet I could guess the factorization faster than the reverse foil method. So I'm going to try and guess and check here. The only way you could have a 2 sine squared would be you have a 2 sine of 3 theta right here and you have just a sine of 3 theta. Notice 2 sine times sine gives you 2 sine squared. Also, the only way you could get a 1 is a 1 times 1 if we just think of integer factorizations. Now, because this is a negative sign right here, 1 has to be positive. 1 has to be negative. And I'm going to put the negative here in the plus here. Notice, of course, 1 times negative 1 is negative 1. And we get 2 sine times negative 1. That gives us a negative 2 sine. And then we have a plus 1 sine that adds together to give us a negative 1. So that is the correct factorization. This is equal to 0. So then by the 0 product property, we can set each factor equal to 0 and solve it from there. So consider the first one. 2 sine of 3 theta plus 1. This becomes 2 sine 3 theta equals negative 1. You just track one from both sides. Divide by 2. We're going to get sine of 3 theta equals negative 1 half. So when is sine equal to negative 1 half? This would happen in the third quadrant and in the fourth quadrant. In the first quadrant, the reference here is going to be 30 degrees. So I'm looking for angles in the third and fourth quadrants, which reference the 30 degrees. So we get 3 theta equals. In the third quadrant, we're going to take 180 plus 30 degrees, which is of course going to give us 210 degrees. We got all solutions. So we have to put in the 360K. Of course, also we have multiple solutions. You always have to consider the general period because that will give us other solutions as well. So the solution in the fourth quadrant will of course be 360 take away 30 degrees, which is 330 degrees plus 360K like so. And so then let's move on to the second factor here. Sine 3 theta minus 1 is equal to 0. Well, if you add 1 to both sides, you get sine of 3 theta is equal to 1. And when is sine equal to 1 that happens at the top of the unit circle that happens at 90 degrees plus of course 360K. So we have three general solutions to when sine equals negative 1 half or 1, but we need to divide all of this by 3. So cut everything by a third. Notice this will cut the period by a third as well. We get that theta is going to equal. Let's take 120 and divide it by 3. That's going to give us 70 degrees plus 120 degrees K. We're going to take 330 cut it by a third. That's going to give us 110 degrees plus 120K. And then we have to take a third of 90 degrees, which is 30 degrees plus 120K. That's like so. In which case this gives us the general solution right here. If we wanted individual solutions, we'd plug in K equals 0, K equals 1, K equals 2 and write all nine of those possibilities. Let's look at a quadratic equation involving tangent here. Tangent squared 3x equals 1. Well, this one's just because it's tangent squared makes life a lot easier. We're just going to take the square root of both sides, just the square root here. But when you take the square root, of course, you have to consider two possibilities. It's the square root of 1 and negative 1 here. I'm going to break it off in these two cases. There's tangent of 3x equals 1. And then there's tangent of 3x equals negative 1. When does tangent equal positive 1 that happens in the first quadrant at 45 degrees? It does ask for radian solutions. That's going to happen at pi force. So 3x is equal to pi force plus pi K. Tangent and cotangent are pi periodic, so their standard period is just pi. Of course, when you take a third of that, you're going to end up with x equals pi 12 plus pi thirds K. Like so. What about the second one? What is tangent equal to negative 1? That happens, of course, in the second and fourth quadrant. 3x here, you're going to get 3 pi force plus pi K, like so. Divide both sides by 3 and you'll end up with pi force plus pi thirds K. So I took 3 pi force divided by 3. That makes it 3 pi 12. But I took the liberty of simplifying that just to be pi force right there. And so the general solution here is going to be x equals pi 12 and pi force. And then you could add any multiple of pi thirds to that. And that would give you all the solutions in radian measure.