Đặt $t=\sin x-\cos x=\sqrt2\sin\left(x-\dfrac{\pi}{4}\right)$ ($t\in [-\sqrt2;\sqrt2]$)
$\to t^2=1-2\sin x\cos x$
$\to \sin x\cos x=\dfrac{1-t^2}{2}$
PT trở thành:
$\dfrac{1-t^2}{2}=6(t-1)$
$\to 1-t^2=12t-12$
$\to t^2+12t-13=0$
$\to t=1$ (do $t\in [-\sqrt2;\sqrt2]$)
$\to \sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt2}$
$\to \left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi \\x=\pi+k2\pi \end{array} \right.$
