Trying to solve equation $\sin^2 x = 1$ The assignment is that I'm suppose to correct a flawed solution to the equation $\sin^2 x = 1$.
The flawed solution is: 
$\sin^2 x = 1$
$\sin x = 1$
$x = 90^\circ + 2n\pi$
I thought I was simply to correct the fact that they forgot the negative root. So my solution was, skipping the solving steps here obviously: 
$x = 2n\pi - 90^\circ$
However my teacher says that I am repeating an error that was also present in the flawed solution. Apparently it had something to do with the last line in the original solution. I have also stated that $n$ is an arbitrary integer.
Any help is greatly appreciated!
EDIT:
I think the error the teacher is referring to is that I'm expressing the solution a bit weird by mixing radians and degrees, like you guys mentioned. I've also defined both roots more clearly like @amWhy did below. 
Thanks guys!  
 A: $\cos^2x=1-\sin^2x=0\implies \cos x=0\implies x=\frac{(2n+1)\pi}2$ where $n$ is any integer

Alternatively,
$\sin^2x=1\implies \cos2x=1-2\sin^2x=1-2\cdot 1=-1=\cos\pi$
So,$2x=2n\pi+\pi=(2n+1)\pi$ where $n$ is any integer
So, $x=\frac{(2n+1)\pi}2$
A: Hint: Your solution omitted what the original solution conveyed, and the original solution omitted what your solution conveyed:
You need to find solutions for both roots of $\sin^2 x = 1$:
$$\sin^2 x = 1 \iff \sin x = 1 \;\;\text {or} \;\;\sin x = -1$$
Any $x$ satisfying either of the equations is a solution to the given equuation.

Also note that both the original solution and your solution: $\pm 90^\circ + 2n\pi\;$ mixes degrees with radians; by convention, we express the solutions with the consistent use of units.
$$x = 2n\pi \pm \frac{\pi}{2}\quad n \in \mathbb Z \quad\quad \text{or, equivalently} \quad \quad x= {k\pi} + \pi/2 \quad k \in \mathbb Z $$ 
A: One way to avoid the $\pm 1$ issue is to notice that $\cos^2 x + \sin^2 x = 1$. Then, solutions exist when $\cos^2 x = 0$, which is exactly when $\cos x = 0$.
