How to solve equations like $\alpha \sin x -\beta\sin 2x +\gamma=0 $

Question

Can I solve this equation without Newton-Raphson method? I have $\alpha=47.02$ $\beta=112.5$ and $\gamma=50$. When I have to use Newton-Rapson to solve trigonometric equations ? I will greatly appreciate your answers.

Answer 1

Hint: Substitute $$\sin(x)=\frac{2t}{1+t^2}$$

$$\cos(x)=\frac{1-t^2}{1+t^2}$$ the so called Weierstrass substitution.

Answer 2

Alt. hint:   write it as $\, 2 \beta \sin x \cos x = \alpha \sin x + \gamma\,$, and square both sides. With $\,s = \sin x\,$ the equation then becomes a depressed quartic: $\;4 \beta^2 s^2(1-s^2) = (\alpha s + \gamma)^2\,$.