 In this video, we provide the solution to question number 25 from the practice final exam for math 1060 In which case we have to solve the equation for cosine theta minus 3 secant theta equals 0 And we're going to do this in radians only find the solutions from 0 to 2 pi where 0 actually is included in that So notice I have different trigonometric functions here cosine and secant if I want to solve this one I would love it to be the same trigonometric function now I noticed that secant of course is just 1 over cosine so you can get 4 cosine theta minus 3 over cosine theta is equal to 0 like so so now it's just cosine which is great But I want to solve this equation. It's gonna be hard to do with cosines of the denominator So I need times both sides of the equation by cosine theta So we're gonna clear the denominators the nice thing about the right hand side being zeros when you times by cosine It's still 0 so on the right hand side you get 0 on the left hand side This is we have to be careful you're gonna distribute this cosine on to each piece you get cosine times 4 cosine Which is 4 cosine squared and then you're gonna get negative 3 over cosine times cosine those cosines would cancel out And you end up with just a negative 3 like so solving for cosine. We're gonna add 3 to both sides This gives us 4 cosine squared theta equals 3 divide both sides by 4 So we end up with cosine squared theta is equal to 3 fourths and now take the square root of both sides Now notice when you take the square root you have to consider positive and negative There's two possibilities there So we get cosine theta is equal to plus or minus the square root of 3 over 4 now This is probably not the one you're used to we really should think of it as a square root of 3 over square to 4 Which the square root of 4 is 2 so plus or minus square 3 over 2 aha You're probably familiar with that ratio So when is cosine equal to root 3 over 2 we're doing this in radians remember so in the first quadrant Well, I mean this is actually so cosine is positive in the first and fourth quadrant But it's negative in the second and third quadrant. So it turns out we're gonna grab all of them In the first quadrant, this would be 30 degrees or in other words Pi 6 like so in the second quadrant. This would occur at 5 pi 6th in The third quadrant this happens at 7 pi 6th and then the fourth quadrant you end up with 11 pi 6th And so that's what those would give you the four solutions to this trigonometric equation