# 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?

• You can reduce the problem to 4-degree polynom by replacing $y=sin(x)$, $\pm\sqrt{1-y^2} = cos(x)$. – hvedrung Aug 20 '15 at 12:43
• It is not at all a so simple equation, be sure. For the general case, numerical methods would be required. – Claude Leibovici Aug 20 '15 at 12:46
• equivalent form is $\sin{\gamma}\cdot t^4-2(2a+\cos{\gamma})t^3+2(2a-\cos{\gamma})t-\sin{\gamma}=0,t=\tan{\dfrac x 2}$ which can be solved ,but solutions are very uggly. – Booldy Aug 20 '15 at 13:23
• Thanks. If it reduces to a function to degree 4, then a numeric approximation is probably faster than calculating the solution(s) directly... – fishinear Aug 20 '15 at 16:04

## 1 Answer

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 $$a^2 = \frac{3\sin(x + \gamma)^2}{4} + \frac{1}{4}.$$

• There is no point in differentiating both sides, this is equation not identity. – user261263 Aug 20 '15 at 13:13
• I forgot that. My bad. – exilednick Aug 20 '15 at 13:16