tìm tất cả các nghiệm của phương trình sinxcosxcos2xcos4xcos8x= $\frac{1}{16}$ sin12x với x ∈ [- $\frac{pi}{2}$; $\frac{\pi}{2}$]

$\begin{array}{l} \sin x\cos x\cos 2x\cos 4x\cos 8x = \frac{1}{{16}}\sin 12x\\ \Leftrightarrow 16\sin x\cos x\cos 2x\cos 4x\cos 8x = \sin 12x\\ \Leftrightarrow 8\sin 2x\cos 2x\cos 4x\cos 8x = \sin 12x\\ \Leftrightarrow 4\sin 4x\cos 4x\cos 8x = \sin 12x\\ \Leftrightarrow 2\sin 8x\cos 8x = \sin 12x\\ \Leftrightarrow \sin 16x = \sin 12x\\ \Leftrightarrow \left[ \begin{array}{l} 16x = 12x + k2\pi \\ 16x = \pi - 12x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 4x = k2\pi \\ 28x = \pi + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{k\pi }}{2}\\ x = \frac{\pi }{{28}} + \frac{{k\pi }}{{14}} \end{array} \right.\\ TH1:x = \frac{{k\pi }}{2}\\ x \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right] \Rightarrow - \frac{\pi }{2} \le \frac{{k\pi }}{2} \le \frac{\pi }{2} \Leftrightarrow - 1 \le k \le 1 \Rightarrow k \in \left\{ { - 1;0;1} \right\}\\ \Rightarrow x = - \frac{\pi }{2};0;\frac{\pi }{2}\\ TH2:x = \frac{\pi }{{28}} + \frac{{k\pi }}{{14}}\\ x \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right] \Rightarrow - \frac{\pi }{2} \le \frac{\pi }{{28}} + \frac{{k\pi }}{{14}} \le \frac{\pi }{2} \Leftrightarrow - \frac{{15}}{2} \le k \le \frac{{13}}{2} \Rightarrow k \in \left\{ { - 7; - 6;...;6;7} \right\}\\ x = - \frac{{13\pi }}{{28}}; - \frac{{11\pi }}{{28}}; - \frac{{9\pi }}{{28}}; - \frac{{7\pi }}{{28}}; - \frac{{5\pi }}{{28}}; - \frac{{3\pi }}{{28}}; - \frac{\pi }{{28}};\frac{\pi }{{28}};\frac{{3\pi }}{{28}};\frac{{5\pi }}{{28}};\frac{{7\pi }}{{28}};\frac{{9\pi }}{{28}};\frac{{11\pi }}{{28}};\frac{{13\pi }}{{28}} \end{array}$