I have the equation:
$$\alpha(\sin\beta x \cos\beta x - \cos \beta x \sin \beta x) = 1$$
I can't seem to get this in the form where it's suitable for the $\sin(2\alpha)$ identity, however this seems like an interesting final solution. I was wondering if this equivalency meant there are any special conditions on alpha and beta for this to work? I am struggling to formulate any. Thanks!
Using the angle sum/difference identity, you have $$ \alpha (\sin(0))=1 \implies 0 = 1 \to\gets $$
a contradiction. Changing the right-hand side to $0$, the equation is satisfied for all $\alpha,\beta \in \mathbb{R},\mathbb{C}$. This solution set is trivial.
Swapping $-$ with $+$ gives us $$ \alpha \sin(2\beta x) =1 \implies 2\alpha \sin(\beta x)\cos(\beta x) = 1 $$ which gives the solutions $$ x = \frac{\sin^{-1}(1/\alpha) + 2n\pi}{2\beta},\frac{\pi -\sin^{-1}(1/\alpha) + 2n\pi}{2\beta};\,\alpha,\beta \neq 0 \text{ and } n \in \mathbb{Z}\, \text{.} $$