<s>[INST] 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?

 [/INST] $\sin(2a-45) = \frac{\sqrt{3}}{2}$ 
$2a-45 = 60 + 360n \Rightarrow a = 52.5  + 180n$ 
$2a-45 = 120+ 360n \Rightarrow a = 82.5 + 180n$
Either $a = 52.5 $ or $a = 82.5$. If you square 
$\sin(4a) = \frac{-1}{2}$ 
$4a = 210 + 360n \Rightarrow a = 52.5 + 90n $ 
$4a = 330 + 360n \Rightarrow a = 82.5 + 90n $ 
First two solution
$a = 52.5 $ 
$a= 52.5 - 90 = -38.5$ 
The other two
$a = 82.5 , 82.5 - 90 = -7.5$ 
In both cases the second solution is rejected see 
Checking $\sin(-15) - \cos(-15) = \frac{-\sqrt{6}}{2}$ which is wrong same for the other
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