Đặt \(\left|sinx-cosx\right|=t\Rightarrow\left\{{}\begin{matrix}0\le t\le\sqrt{2}\\sin2x=2sinx.cosx=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(1-t^2-2mt-1+3m^2=0\)
\(\Leftrightarrow t^2+2mt-3m^2=0\)
\(\Leftrightarrow\left(t-m\right)\left(t+3m\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=m\\t=-3m\end{matrix}\right.\)
Để pt có nghiệm \(\Leftrightarrow\left[{}\begin{matrix}0\le m\le\sqrt{2}\\0\le-3m\le\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0\le m\le\sqrt{2}\\-\frac{\sqrt{2}}{3}\le m\le0\end{matrix}\right.\) \(\Rightarrow-\frac{\sqrt{2}}{3}\le m\le\sqrt{2}\)
