How to solve $\sin{14x} - \sin{12x} + 8 \sin x - \cos{13x} = 4$? Find all solutions of $$\sin{14x} - \sin{12x} + 8 \sin x - \cos{13x} = 4$$ on an interval $(0^\circ, 360^\circ)$

Using the trigonometric identity $$\sin A-\sin B\equiv2\cos\frac{A+B}2\sin\frac{A-B}2,$$with $A=14x$ and $B=12x$, we can write the equation as $$(2\sin x-1)(\cos13x+4)=0.$$The general solution is therefore $$x=n\pi+(-1)^n\frac\pi6\quad(n\in\Bbb Z).$$For the range you specify, $x$ is $30^\circ$ or $150^\circ$. 
