<s>[INST] 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. 
 [/INST] 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.
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