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Điều kiện: $\sin 2x\ne0$
$\frac{1}{\sin x}+\frac{1}{\displaystyle \sin(x-\frac{3\pi}{2} )}=4\sin (\frac{7\pi}{4}-x )$
$\Leftrightarrow
\frac{1}{\sin x}+\frac{1}{\displaystyle \sin(x+\frac{\pi}{2} )}=4\sin (\frac{-\pi}{4}-x ) $
$\Leftrightarrow \frac{1}{\sin x}+\frac{1}{\displaystyle \cos x}=-4\sin (x+\frac{\pi}{4})$
$\Leftrightarrow \frac{\displaystyle 2\sqrt2\sin(x+\frac{\pi}{4})}{\sin2x}=
-4\sin (x+\frac{\pi}{4})$
$\Leftrightarrow \left [ \begin{array}{l} \sin (x+\frac{\pi}{4})=0\\ \sin2x=\frac{-1}{\sqrt2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x+\frac{\pi}{4}=k\pi\\ 2x=\frac{-\pi}{4}+2k\pi\\2x=\frac{5\pi}{4}+2k\pi \end{array} \right. (k\in\mathbb{Z})$
$\Leftrightarrow
\left[ \begin{array}{l} x=\frac{-\pi}{4}+k\pi\\ x=\frac{-\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi \end{array} \right. (k\in\mathbb{Z}) $
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