Using the R method for finding all solutions to $\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$. My solution differs from official answer. 
How many solutions does
  $$\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$$ 
  have between $-90^\circ$ and $90^\circ$? 

I used the R method and got
$$2a-45^\circ = \arcsin\left(\frac{\sqrt{3}}{2}\right).$$ 
Since $a$ is between $-90^\circ$ and $90^\circ$, then $2a$ is between $-180^\circ$ and $180^\circ$. The RHS can be $60^\circ$, $120^\circ$, $-240^\circ$, and $-300^\circ$. Only $60^\circ$ and $120^\circ$ fit the criteria, but the answer is 4 solutions. 

Where did I go wrong?


Hint:
$$2a-45^\circ=180^\circ n+(-1)^n\arcsin\dfrac{\sqrt3}2$$
If $n$ is odd$=2m+1$(say)
$$2a-45=360m+180-60$$
But $-180-45\le2a-45\le180-45$
$-225\le360m+120\le135$
$?\le m\le?$
What if $n$ is even $=2m$(say)
