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Điều kiện: $\left\{\begin{array}{l}\sin2x\ne0\\\cot x\ne-1\end{array}\right.\Leftrightarrow \left\{\begin{array}{l}x\ne k\dfrac{\pi}{2}\\x\ne\dfrac{-\pi}{4}+k\pi\end{array}\right.,k\in\mathbb{Z}$
Phương trình đã cho tương đương với:
$\dfrac{1+\dfrac{\sin x}{\cos x}}{1+\dfrac{\cos x}{\sin x}}=2\sin x$
$\Leftrightarrow \dfrac{\sin x}{\cos x}=2\sin x$
$\Leftrightarrow \sin2x=\sin x$
$\Leftrightarrow \left[\begin{array}{l}2x=x+k2\pi\\2x=\pi-x+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=k2\pi\\x=\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\end{array}\right.,k\in\mathbb{Z}$
Kết hợp đk, ta có: $x\in\{\dfrac{\pi}{3}+k2\pi;\dfrac{-\pi}{3}+k2\pi,k\in\mathbb{Z}\}$
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