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c.
Phương trình đã cho tương đương với:
$8\cos^3x-2(2\cos^2x-1)-4\cos x-1=0$
$\Leftrightarrow 8\cos^3x-4\cos^2x-4\cos x+1=0$
$\Rightarrow (\cos x+1)(8\cos^3x-4\cos^2x-4\cos x+1)=0$
$\Leftrightarrow 8\cos^4x-8\cos^2x+1+4\cos^3x-3\cos x=0$
$\Leftrightarrow \cos4x+\cos3x=0$
$\Leftrightarrow \cos4x=\cos(3x-\pi)$
$\Leftrightarrow \left[\begin{array}{l}4x=3x-\pi+k2\pi\\4x=-3x+\pi+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=-\pi+k2\pi\\x=\dfrac{\pi}{7}+k\dfrac{2\pi}{7}\end{array}\right.,k\in\mathbb{Z}$.
Thử lại, ta được: $x\in\{\dfrac{\pi}{7}+k2\pi;\dfrac{-\pi}{7}+k2\pi;\dfrac{3\pi}{7}+k2\pi;\dfrac{-3\pi}{7}+k2\pi;\dfrac{5\pi}{7}+k2\pi;\dfrac{-5\pi}{7}+k2\pi,k\in\mathbb{Z}\}$
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