Is $\sin^2\theta + \cos\theta = 2$ solvable without a mess? 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? 
 A: 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 $\log$).
A: The equation has no real solutions.
For every $\theta\in\mathbb R$, we have $\sin^2\theta\in[0,1]$ and $\cos\theta\in[-1,1]$. This means that $\sin^2\theta+\cos\theta=2$ is only possible if $\sin^2\theta=1$ and $\cos\theta=1$. But if $\sin^2\theta=1$ we immediately have $\cos^2\theta=1-\sin^2\theta=0$, so $\cos\theta$ would have to be equal to $0$. This means the equation has no real solutions.
A: we have $|\sin[x]|\le 1$, thus $|\sin[x]|^{2}\le 1$. Your equation would imply both $\sin[x]$ and $\cos[x]$ has absolute value 1, which does not hold since then $\sin[2x]=2\sin[x]\cos[x]=\pm2$. Maybe you copied the wrong formula, etc. 
A: Because cosine is a wave that has a domain of all real numbers, you can plug in multiple different x's for the same y's. When you're solving this, remember that cosθ can give you different values.
Start by using the Pythagorean identity (which you did) and get cos^2θ−cosθ+1=2 (which is slightly off on the post). Move the 1 to the left, then factor out cosine: -cosθ(cosθ-1)=1
Now, set both parts to equal 1:
-cosθ=1
cosθ-1=1
you get that cosθ = -1 and cosθ = 0
Now, use the inverse to find what θ is!
cos^-1(0) = θ
and
cos^-1(-1) = θ
I got pi and pi/2, hopefully that's right!
