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Điều kiện: $\cos x\ne0 \Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi,k\in\mathbb{Z}$
Phương trình đã cho tương đương với:
$\cos2x+\dfrac{2\cos x\sin^2x}{\sin^2x}-\cos x=2$
$\Leftrightarrow 2\cos^2x-1+\dfrac{2\sin^2x}{\cos x}-\cos x=2$
$\Leftrightarrow 2\cos^3x-\cos x+2(1-\cos^2x)-\cos^2x=2\cos x$
$\Leftrightarrow 2\cos^3x-3\cos^2x-3\cos x+2=0$
$\Leftrightarrow \left[\begin{array}{l}\cos x=-1\\\cos x=\dfrac{1}{2}\end{array}\right.$, vì $|\cos x|\le1$
$\Leftrightarrow \left[\begin{array}{l}x=\pi+k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{array}\right.,k\in\mathbb{Z}$
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