Hey everyone. In this problem, we're asked to find all solutions to the equation the tangent of theta is equal to the square root of 3. So, remember the first thing we want to do here is find all the angles for which this is true on our unit circle. Then we can proceed with adding 2πn to each of those solutions. So, starting on my unit circle, I want to find all the values for which my tangent is root 3. So in quadrant 1, I know that for my angle π3, my tangent is equal to the square root of 3. Now also knowing that in quadrant 3, all of my tangent values are positive here and I'm looking for positive square root of 3, I know that from my angle 4π3, my tangent is also equal to the square root of 3. So here, two answers that I get are π3 and also 4π3. Now we can just go ahead and add 2πn to each of these solutions in order to get all of the solutions to this equation. So if we take both of these, add 2πn, this represents all of the solutions to our equation. But something that you actually might notice here is that π3 and 4π3 are exactly π radians away from each other. So, if I were to just take π3 and add π to it, I would still get all of my solutions. So starting with π3, if I add one rotation of π, if I add another one and then another one and then another one, I will still get all of my solutions. So while π3+2πn and 4π3+2πn still represent all of my solutions, if I want to further simplify this, I can write this as π3+πn. And this represents the most simplified form of all of my solutions because when working with the tangent, if I have two angles for which this is true, they're actually always going to be π radians away from each other. So this doesn't happen for every single trig function or trig equation that you work with. But you may notice that if your angles are exactly π radians away from each other, you can actually simplify this further. So our final answer here, in its most simplified form, is π3+πn for the tangent of theta being equal to the square root of 3. Thanks for watching, and I'll see you in the next one.