Using the R method for finding all solutions to $\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$. My solution differs from official answer.

Question

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?

Answer 1

You are right. Here is the image of the function using google.

Answer 2

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)

Answer 3

$\sin(2a-45) = \frac{\sqrt{3}}{2}$

$2a-45 = 60 + 360n \Rightarrow a = 52.5 + 180n$

$2a-45 = 120+ 360n \Rightarrow a = 82.5 + 180n$

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