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Phương trình đã cho tương đương với:
$(\sin x-\cos x)(\sin^2x+\sin x\cos x+\cos^2x)=3\sin x\cos x-1$
$\Leftrightarrow (\sin x-\cos x)(1+\sin x\cos x)=3\sin x\cos x-1$
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:
$t(1+\dfrac{1-t^2}{2})=3\dfrac{1-t^2}{2}-1$
$\Leftrightarrow t^3-3t^2-3t+1=0$
$\Leftrightarrow \left[\begin{array}{l}t=-1\\t=2-\sqrt3\end{array}\right.$, vì $|t|\le\sqrt2$
Khi đó:
$\left[\begin{array}{l}\sin(x-\dfrac{\pi}{4})=\dfrac{-1}{\sqrt2}\\\sin(x-\dfrac{\pi}{4})=\dfrac{2-\sqrt3}{\sqrt2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=k2\pi\\x=\dfrac{3\pi}{2}+k2\pi\\x=\arcsin\dfrac{2-\sqrt3}{\sqrt2}+\dfrac{\pi}{4}+k2\pi\\x=\pi-\arcsin\dfrac{2-\sqrt3}{\sqrt2}+\dfrac{\pi}{4}+k2\pi\end{array}\right.,k\in\mathbb{Z}$
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