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Đặt: $t=\sin x-\cos x=\sqrt2\sin(x-\dfrac{\pi}{4}),|t|\le\sqrt2$, khi đó: $\sin x\cos x=\dfrac{1-t^2}{2}$
Phương trình trở thành:
$t=2\sqrt6\dfrac{1-t^2}{2}$
$\Leftrightarrow \sqrt6t^2+t-\sqrt6=0$
$\Leftrightarrow \left[\begin{array}{l}t=\dfrac{2}{\sqrt6}\\t=\dfrac{-\sqrt6}{2}\end{array}\right.$
Khi đó:
$\left[\begin{array}{l}\sin(x-\dfrac{\pi}{4})=\dfrac{1}{\sqrt3}\\\sin(x-\dfrac{\pi}{4})=\dfrac{-\sqrt3}{2}\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\arcsin\dfrac{1}{\sqrt3}+\dfrac{\pi}{4}+k2\pi\\x=\pi-\arcsin\dfrac{1}{\sqrt3}+\dfrac{\pi}{4}+k2\pi\\x=\dfrac{-\pi}{12}+k2\pi\\x=\dfrac{19\pi}{12}+k2\pi\end{array}\right.,k\in\mathbb{Z}$
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