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!  

$\cos^2x=1-\sin^2x=0\implies \cos x=0\implies x=\frac{(2n+1)\pi}2$ 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, $x=\frac{(2n+1)\pi}2$
