TITLE: Is $\sin^2\theta + \cos\theta = 2$ solvable without a mess?
QUESTION [7 upvotes]: I'm given the problem $\sin^2\theta + \cos\theta = 2$ and I'm told to use the pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to solve it.
I end up with $\cos^2\theta - \cos\theta + 1 = 0$, but I know that's not going to factor and solve very nicely. 
Did I do something wrong, or is the answer going to end up being very ugly?

REPLY [5 votes]: Almost there.  Solve the quadratic equation and get $\cos\theta={1\pm \imath\sqrt{3}\over 2}$.  Take $\pm\cos^{-1}({1\pm \imath\sqrt{3}\over 2})$ to get the answer.  (It's $\pm$ since both $\cos$ and $\sin^2$ are even functions.)  You can look up how to do complex $\cos^{-1}$ and $\log$ (you'll see why you need $\log$).