Hey everyone. In working with trig functions, you've likely come across a statement such as this one, the sine of theta equals 1 half. But this statement is actually more explicitly a trig equation, which we're going to be diving deeper into now. Now in working with the unit circle, you likely also know that there are multiple angles for which this is true. The sine is equal to 1 half, which leads us to the conclusion that trig equations have multiple solutions. But how exactly can we go about finding all of these solutions? Well, here I'm going to walk you through exactly how to find all of the solutions to trig equations such as this one just using the unit circle and adding in one extra thing. So let's go ahead and get started here.

Now in our example here, we're asked to find all of the solutions to the equation within the interval 0 to 2 pi, which is just one rotation around our unit circle. So here we want to find all of the angles for which the sine is equal to 1 half. So coming to our unit circle over here starting in that first quadrant, I know that for my angle pi over 6, the sine is equal to 1 half. So that represents a solution to my trig equation here, θ=π6. Now continuing around my unit circle going into quadrant 2, I know that for 5pi over 6, the sine is also equal to 1 half. So this represents yet another solution to my trig equation. Now in quadrants 3 and 4, I know that my sine values are going to be negative. So there are no other solutions in here. So pi over 6 and 5 pi over 6 represent the solutions to my trig equation within the interval 0 to 2 pi.

But often you're not just going to be limited to that interval 0 to 2 pi, and you're actually going to be asked to find all of the solutions to the equation with no restrictions. So what exactly does that mean to find all of the solutions to a trig equation? Well, for my trig equation here, the sine of theta equals 1 half, I already know that pi over 6 and 5 pi over 6 represent 2 of the solutions to this equation. But if I were to continue another rotation around my unit circle and come back to what would now be 13 pi over 6, the sine of 13 pi over 6 is also equal to 1 half. So that represents yet another solution to this equation. And then if I were to continue on for another rotation and reach what would now be 17 pi over 6, the sine of 17pi over 6 is also equal to 1 half. So this represents yet another solution. And if I were to continue going around and around and around my unit circle, I would have an infinite number of solutions. So how do we account for all of those solutions? Well, you may be worried that it's going to be a bit tricky, but it's actually rather simple because all we're going to do is first find all of the solutions on our unit circle and then simply add 2 pi n to each of them, where 2 pi represents a full rotation around my unit circle and n is just an integer that represents how many times you're going around it. Now you might also see this written as 2 pi k depending on your professor or your textbook, but it means the same exact thing. So let's look at our sineθ=1/2 and identify all of the solutions using this method.

Now we first want to find all of those solutions on our unit circle, which we already did in our earlier example. So we know that on our unit circle, we have our 2 solutions of pi over 6 and 5 pi over 6 that make this statement true. The sine of it is equal to 1 half. Now that we have these solutions that are on our unit circle, all that's left to do is add 2 pi n to each of them. And we've now covered all of our bases. This represents all of the possible solutions to this equation, the sine of theta equals one half, and we're done here. This is our final answer.

Now let's take a look at one other example here. Here we're asked to find all of the solutions to the equation that the cosine of x equals negative one. Now the first thing you might notice here is that we're dealing with an x rather than a theta, but that doesn't change anything. We're still trying to find the angles that make this equation true. So coming up to our unit circle, we first want to find all of our solutions that are already on there and then go from there. So we want to find all of the solutions for which our cosine is equal to negative one. So on our unit circle here, if I go around looking for my cosine values of negative one, there's actually only one angle for which this is true and that's pi. So my solution that is on my unit circle is x equals pi. Now all that's left to do is add 2 pi n to this pi and we're done. This represents all of the possible solutions to the equation the cosine of x equals negative one, and we're good to go. Now that we know how to find all of the solutions to basic trig equations, let's get some more practice together. Thanks for watching, and let me know if you have any questions.