Solving $\cos^2{\theta}-\sin{\theta} = 1$ Can someone please help me solve this?

$$\cos^{2}{\theta}-\sin{\theta} = 1, \quad\theta\in[0^{\circ}, 360^{\circ}]$$

 A: Notice, we have 
$$\cos^2\theta-\sin\theta=1$$
$$1-\sin^2\theta-\sin\theta=1$$
$$-\sin^2\theta-\sin\theta=0$$
$$\sin^2\theta+\sin\theta=0$$
$$\sin\theta(\sin\theta+1)=0$$ $$\sin\theta=0\iff \theta=n(180^\circ)$$
Where, $n$ is any integer. 
But for the given interval $[0^\circ, 360^\circ]$, substituting $n=0, 1, 2$, we get 
$$ \theta=0^\circ, 180^\circ, 360^\circ$$
Now, $$\sin\theta+1=0$$
$$\sin\theta=-1\iff \theta=2n(180^\circ)-90^\circ$$
But for given interval $[0^\circ, 360^\circ]$, substituting $n=1$ we get
$$\theta= 270^\circ$$
Hence, we have $$\color{red}{\theta}=\left\{\color{blue}{0^\circ, 180^\circ, 270^\circ, 360^\circ} \right\}$$
A: $\cos^{2}(\theta) - \sin(\theta) = \cos^{2}(\theta) + \sin^{2}(\theta)$, (using Pythagorean Identity)
$\sin^{2}(\theta) + \sin(\theta) = 0$,  
let $x = \sin(\theta)$
$x^{2}+ x = 0$
$x=\frac{-1 \pm \sqrt{1^{2}- 4(1)(0)}}{2}$, (Quadratic Formula)
$x = 0$, $x = -1$ 
substitute $\sin(\theta)$ back in and solve
$0 = \sin(\theta) \Rightarrow \theta = 0^{\circ},180^{\circ},360^{\circ}$
$-1 = \sin(\theta) \Rightarrow \theta = -270^{\circ}$
$\theta = \{0^{\circ},180^{\circ},270^{\circ},360^{\circ}\}$.
