Solve trigonometric equation $\sin14x - \sin12x + 8\sin x - \cos13x= 4$ I am trying to solve the trigonometric equation 
$$ \sin14x - \sin12x + 8\sin x - \cos13x= 4 $$ 
The exact task is to find the number of real solutions for this equation on the range $[0, 2\pi]$. Thanks. 
 A: We can use $\sin{x}-\sin{y}=2\sin{\frac{x-y}{2}}\cos{\frac{x+y}{2}}$.
So, $$(\sin{14x}−\sin{12x})+8\sin{x}−\cos13x=4 \Longleftrightarrow 2\sin{x}\cos{13x}−\cos13x+8\sin{x}-4=0$$ $$\Longleftrightarrow 2\cos{13x}(\sin{x}-\frac{1}{2})+8(\sin{x}-\frac{1}{2})=0 \Longleftrightarrow (\sin{x}-\frac{1}{2})(2\cos{13x}+8)=0$$
So, we have $\sin{x}=\frac{1}{2}$ and $\cos{13x}=-4$(which can't be achieved), now we take just $\sin{x}=\frac{1}{2}$, and we get $x_1=\frac{\pi}{6}$, $x_2=\frac{5\pi}{6}$ on the given interval.
A: Define:
$$\sin(14x) - \sin(12x) + 8\sin x - \cos(13x)- 4=:F(\cos x,\sin x)$$
Then set $t=\tan(x/2)$, we have
$$\cos x=\frac{1-t^2}{1+t^2}, \sin x=\frac{2t}{1+t^2}$$
$$G(t)=F\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)$$
$$=-\frac{t^2-4t+1}{(1+t^2)^{14}}H(t)$$
$$H(t)=5 - 273t^2 + 15262t^4 - 229086t^6 + 
   1565135t^8 - 5306587t^{10} + 
   9664564t^{12} - 9650836t^{14} + 
   5316883t^{16} - 1559415t^{18} + 
   231374t^{20} - 14638t^{22} + 377t^{24} + 
   3t^{26}$$
Numerical results showed that all the roots of $H(t)$ are non real. So the real roots of  $G(t)$ are those from 
$$t^2-4t+1=0$$
Therefore we have
$$t_1=2-\sqrt{3},t_2=2+\sqrt{3}$$
$$x_1=\frac{\pi}{6},x_2=\frac{5\pi}{6}$$
