Solving $\sin^2 \theta +\cos \theta=1$ Can someone please explain where the $1$ goes in this expression?

Find the measure of angle in radians by solving: 
  $$\sin^2 \theta +\cos \theta=1$$

I got $$\sin^2 \theta = 1- \cos \theta$$
Substituting, I get:
$$(1-\cos \theta) +\cos \theta -1= 0$$
I know the next step is $$\cos (1-\cos) = 0$$
But where does the $1$ go? Does it cancel out and if so why?
 A: Given $$\sin^2 \theta + \cos \theta = 1,$$ and using the circular identity $$\sin^2 \theta + \cos^2 \theta = 1,$$ it follows that $$\cos^2 \theta = \cos \theta,$$ or $$(\cos \theta - 1) \cos \theta = 0.$$  Hence $$\cos \theta \in \{0, 1\},$$ which implies $$\theta \in \{\tfrac{(2k+1)}{2}\pi, 2k\pi\}, \quad k \in \mathbb Z.$$
You can check this:  if $\theta$ is an integer multiple of $2\pi$, then $\sin \theta = 0$ and $\cos \theta = 1$, so $\sin^2 \theta + \cos \theta = 1$.  If $\theta$ is an odd multiple of $\pi/2$, then $\sin \theta = \pm 1$, so $\sin^2 \theta = 1$, and $\cos \theta = 0$, which also checks out.
A: There's quite a bit going on here.  First, this is not an "identity", it's a conditional equation and you're asked to solve it.  Second, replace $\sin^2 \theta$ by $1-\cos^2 \theta$ (you forgot to square the $\cos \theta.$)  Third, the last time you write $\cos$, you leave it dis-embodied.  You need to take the $\cos$ of something.  (In this case, $\theta.$)
Then you should have something like:
$$1-\cos^2 \theta = 1-\cos \theta.$$
The left side factors by difference of squares.  Then either $1-\cos \theta =0$
(and you solve that equation.) Or you can divide by $1-\cos \theta$ and you can solve the resulting equation.
