Use a graph to solve the given equation for -2π less than or equal to X less than or equal to 2π where eight tangent X equal -8. So if we look at our tangent graph, actually if we look at a graph of Y equal 8 tangent X. So we have our tangent graph that looks something like this, and we're going to have asymptotes π halves, negative π halves, three Pi halves -3 Pi halves. That would be our -2π location, and here would be our 2π location. So if it's an 8IN front, it tells us that we know it's stretched by 8. So somewhere in here about halfway, instead of being at one, it would be now at 8. So we're going to have this be about 8 and this be about -8. These are approximates, obviously, so we want to know when is Y equaling -8. So we want to know when is when are these two graphs intersecting? Well, if we look here, we can see that it's intersecting about here, which would be negative. This is -3 Pi halves, so this would be -5 Pi force. And if we look here, we can see that this is negative π force. And this one would be at three Pi force and this one would be at 7 Pi force. Now if we wanted to just solve it algebraically, we would divide each side by 8 and we'd get tangent X equaling -1, and we know that tangent is -1 Pi force plus 2K Pi and also at 5 Pi force plus 2K Pi. Now the fact that we wanted it in between -2 and two negative 2π and 2π, if K was a -1, we would get π force -2π force or -7 Pi force. If K π force -2 K π force π force -8 Pi is -7 Pi. But that would be in between here. Oh, it's not at π force, OK, It's not at π force, it's at -1. So it's X equal negative, X equal 3 Pi fours plus 2K Pi, and not five Pi fours, but seven Pi fours plus 2K Pi. SO3 Pi fours would be. If K was -1 we'd get -5 Pi fours. If K was 0, we'd get three Pi fours. If K was one, we'd be too big, we'd be past our 2π here. If K is -1, we'd get negative π force, and if K was 0 we'd get 7 Pi force. And if K was two we'd be too big S. Those would be the same values that we found by looking at the graph.