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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.

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    $\begingroup$ I managed to solve this using transform of the first 2 members to product and then obtaining: $ (2sinx-1)(cos13x+4) $ . I realized my little theory doesn't really makes sense and happens to be correct in some specified cases.. $\endgroup$ – Transcendental Jun 17 '14 at 18:19
  • $\begingroup$ After I worked out my solution, I found your solution in the comment. I think that your method is better. This is because f $\cos(13x)+4=0$ has no real solution. So all the real solution are from $2\sin x-1=0$. The solution is, of course $x_1=\pi/6, x_2=5\pi/6$. $\endgroup$ – mike Jun 17 '14 at 18:35
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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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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 $$t_1=2-\sqrt{3},t_2=2+\sqrt{3}$$

$$x_1=\frac{\pi}{6},x_2=\frac{5\pi}{6}$$

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