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