Pt \(\Leftrightarrow6\left(sinx-cosx\right)-sinx.cosx-6=0\)
Đặt \(t=sinx-cosx;t\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(\Rightarrow\dfrac{t^2-1}{2}=-sinx.cosx\)
Pttt:\(6t-\dfrac{t^2-1}{2}-6=0\)
\(\Leftrightarrow-t^2+12t-11=0\Leftrightarrow\left[{}\begin{matrix}t=11\left(ktm\right)\\t=1\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow sinx-cosx=1\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k\pi2\end{matrix}\right.\)\(\left(k\in Z\right)\)
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