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Điều kiện: $-1\le x\le 1$
Đặt $x=\cos t$, với $t\in[0,\pi]$. Dễ thấy $\cos t=x\ne0$.
Phương trình trở thành:
$64\cos^6t-112\cos^4t+56\cos^2t-7=2\sqrt{1-\cos^2t}$
$\Leftrightarrow 64\cos^7t-112\cos^5t+56\cos^3t-7\cos t=2\sin t\cos t$
$\Leftrightarrow \cos 7t=\sin 2t$
$\Leftrightarrow \left[ \begin{array}{l} 7t=\frac{\pi}{2}-2t+2k\pi\\ 7t=2t-\frac{\pi}{2}+2k\pi \end{array} \right. (k\in\mathbb{Z})$
$\Leftrightarrow \left[ \begin{array}{l} t=\frac{\pi}{18}+\frac{2k}{9}\pi\\ t=\frac{-\pi}{10}+\frac{2k}{5}\pi \end{array} \right. (k\in\mathbb{Z})$
$\Rightarrow t\in\{\frac{\pi}{18}, \frac{5\pi}{18} , \frac{13\pi}{18} , \frac{17\pi}{18} , \frac{3\pi}{10} , \frac{7\pi}{10} \}$ vì $t\in[0,\pi]$
Suy ra: $ x\in\{\cos\frac{\pi}{18}, \cos\frac{5\pi}{18} , \cos\frac{13\pi}{18} , \cos\frac{17\pi}{18} ,\cos \frac{3\pi}{10} ,\cos \frac{7\pi}{10} \}$
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