Solving $a \sin 2x = \sin (x + \gamma)$ I am trying to solve the following equation:

$$a \sin 2x = \sin (x + \gamma)$$

or, equivalently:

$$2 a = \frac{\cos \gamma}{\cos x} + \frac{\sin \gamma}{\sin x}$$

where $a$ and $\gamma$ are constants. 
I tried for a long time, and searched the web a lot, but can only find solutions to very simple trigonometry equations, or complex ones where the constants happen to be such that the formula can be reduced to another form. I am surprised that I cannot find a solution to such a simple equation.

Any idea how this can be solved?
Hint: after differentiating both sides, you will get
$$\begin{align}
2a\ cos(2x)=\cos(x + \gamma) \tag{1}\\
a\ sin(2x)=\sin(x + \gamma) \tag{2}\\
\end{align}$$

Squaring and adding $(1)$ and $(2)$, you'll end with
$$a^2 = \frac{3\sin(x + \gamma)^2}{4} + \frac{1}{4}.$$
Hint: after differentiating both sides, you will get
$$\begin{align}
2a\ cos(2x)=\cos(x + \gamma) \tag{1}\\
a\ sin(2x)=\sin(x + \gamma) \tag{2}\\
\end{align}$$

Squaring and adding $(1)$ and $(2)$, you'll end with
$$a^2 = \frac{3\sin(x + \gamma)^2}{4} + \frac{1}{4}.$$