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Điều kiện: $\left\{\begin{array}{l}\cos x\ne0\\\cos^4x\ne0\end{array}\right.\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi,k\in\mathbb{Z}$.
Phương trình đã cho tương đương với:
$\dfrac{\sin^4x+\cos^4x}{\cos^4x}=\dfrac{(2-\sin^22x)\sin3x}{\cos^4x}$
$\Leftrightarrow \sin^4x+\cos^4x=2\sin3x-\sin^22x\sin3x$
$\Leftrightarrow 1-2\sin^2x\cos^2x=2\sin3x-\sin^22x\sin3x$
$\Leftrightarrow 2-\sin^22x=4\sin3x-2\sin^22x\sin3x$
$\Leftrightarrow (2\sin3x-1)(\sin^22x-2)=0$
$\Leftrightarrow \sin3x=\dfrac{1}{2}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{18}+k\dfrac{2\pi}{3}\\x=\dfrac{5\pi}{18}+k\dfrac{2\pi}{3}\end{array}\right.,k\in\mathbb{Z}$, thỏa mãn.
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