 So once you have trigonometric functions, you can also have trigonometric equations. The important thing to remember is that the last operation performed tells us the first thing we must do to solve an equation. So for example, sine 2 theta plus 7 equals 8 is a sum. So we start by subtracting. Square root tangent theta equals 5 is a square root. So we start by squaring. 1 over cosine 2 theta equals 2 is a quotient. So we start by multiplying. And sine of theta equals 1 half. Well, that's actually a trigonometric function. So we start by drawing a picture of the unit circle. So let's try to solve this. Suppose sine of theta equals 1 half. Find the exact values of all possible solutions theta. Now remember, in the context of trigonometric functions, an exact value is an angle expressed in the form p pi over q, where p and q are integers. You might think about these as rational multiples of pi. And these are the exact values you should know. And let's go ahead and try to solve this. The hard way to find all possible solutions is to not draw a picture. But why would you want to do things the hard way? So we'll draw a picture of the unit circle. And remember, on the unit circle, the sine of an angle corresponds to the y value. And since we have sine equals 1 half, we want to find the points where y is equal to 1 half. And so these points are located here and here. Now we see there's two angles, one in the first quadrant and one in the second quadrant. Now remember, there are certain angles whose exact values you should know. And in the first quadrant, we find theta equals pi over 6 has sine of theta equals 1 half. And so this gives us one set of solutions because not only is theta equals pi over 6 1 solution, we can add or subtract any multiple of 2 pi to get other solutions. But wait, there's more. If we want to get to this second solution, we can go halfway around the circle and then back pi over 6. And so this would mean that in the second quadrant, we find pi minus pi over 6, 5 pi over 6, will also have a sine of 1 half. And again, adding or subtracting any multiple of 2 pi will give us additional solutions. Now this approach gives us an infinite number of solutions, but it's sometimes convenient to limit our solutions to those in the interval between 0 and 2 pi. We could also admit solutions where our angle is between 0 and 360 degrees, but we don't admit that. Those are those things, degrees, that we no longer use. But in either case, what we have to do is we limit our values k to those that will put a solution in this interval. Well, how about finding the exact value of all solutions in the interval between 0 and 2 pi to the equation 2 cosine 2 theta equals square root of 2? So the first thing to recognize is the expression involving the unknown is a product. And so we divide by one of the factors. So from 2 cosine 2 theta equals square root of 2, we could divide by 2 cosine 2 to get theta equals square root of 2 over 2 cosine 2. But we would be wrong. And there are three important things to remember. First, cosine theta and sine theta are not products. And so we should never divide by a trigonometric function to solve an equation. Second, cosine theta and sine theta are not products. And so never divide by a trigonometric function to solve an equation. But the most important thing to remember is that cosine and sine are not products. And you should never divide by a trigonometric function to solve an equation. Now, we do still have a product, but it's a product two times some other stuff. And so we'll divide by one of the factors. But since we should never divide by the trigonometric function, we'll have to divide by two. And so we get cosine two theta equals square root of two over two. Since we now have a trigonometric equation, we'll draw a picture. So first, we'll draw a unit circle. Remember that on the unit circle, the cosine corresponds to the x coordinate of a point. So we look for the points where x is square root of two over two. And these are here. I note that we actually have two of them. So again, based on the exact values that we know, we see that in the first quadrant, the angle with the cosine of square root of two over two is pi over four. Since we're taking the cosine of two theta, this means that two theta is pi over four plus or minus any multiple of two pi. Well, that was a waste. We didn't want to find two theta. We wanted to find theta. Fortunately, we can solve for theta by dividing everything by two. Now, since we want to find all solutions in the interval between zero and two pi, we'll choose values of k so that our value theta ends up in this interval. So if k equals zero, theta is pi over eight, which is in the interval between zero and two pi. If k equals one, theta is pi over eight plus pi, nine pi over eight, which is also in this interval. If k equals two, we're taking pi over eight and we're adding two pi, which is guaranteed to put us outside the interval. So we don't need to consider higher values of k. But wait, there's more. Because we drew the picture, we know that there is a second angle that's going to have a cosine of square root of two over two. And that's this angle in the fourth quadrant. And so we see that in the fourth quadrant, the angle with a cosine of square root of two over two is seven pi over four. And so we get solutions of seven pi over four plus or minus any multiple of two pi. But again, this angle is two theta. Since we want to solve for theta, we divide everything by two. And so all solutions will be of the form seven pi over eight plus or minus any multiple of pi. But since we don't want all solutions, we just want all solutions in the interval between zero and two pi. We'll let k take on different values and reject all solutions that fall outside this interval. So if k equals zero, we get seven pi over eight. If k equals one, we get 15 pi over eight. If k equals two, we're trying to add two pi, which will certainly take us outside the interval, so we don't need to worry about k equals two. And so we get four solutions to this trigonometric equation.