\(\Leftrightarrow\frac{4sin2x+cos2x+17}{3cos2x+sin2x+m+1}-2\ge0\) (tất nhiên là với mọi x)
\(\Leftrightarrow\frac{2sin2x-5cos2x-2m+15}{3cos2x+sin2x+m+1}\ge0\)
TH1: \(\left\{{}\begin{matrix}2sin2x-5cos2x-2m+15\ge0\\3cos2x+sin2x+m+1>0\end{matrix}\right.\) ;\(\forall x\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{\sqrt{29}}sin2x-\frac{5}{\sqrt{29}}cos2x\ge\frac{2m-15}{\sqrt{29}}\\\frac{1}{\sqrt{10}}sin2x+\frac{3}{\sqrt{10}}cos2x>\frac{-m-1}{\sqrt{10}}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\left(2x-a\right)\ge\frac{2m-15}{\sqrt{29}}\\sin\left(2x+b\right)>\frac{-m-1}{\sqrt{10}}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2m-15}{\sqrt{29}}\le-1\\\frac{-m-1}{\sqrt{10}}< -1\end{matrix}\right.\) tới đây chắc bạn tự giải được
TH2: tương tự:
\(\left\{{}\begin{matrix}2sin2x-5cos2x-2m+15\le0\\3cos2x+sin2x+m+1< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2m-15}{\sqrt{29}}\ge1\\\frac{-m-1}{\sqrt{10}}>1\end{matrix}\right.\) \(\Leftrightarrow...\)