\(\left(1-m\right)\left(\dfrac{1}{cos^2x}-1\right)-\dfrac{2}{cosx}+1+3m=0\)
\(\Leftrightarrow\left(1-m\right).\dfrac{1}{cos^2x}-\dfrac{2}{cosx}+4m=0\)
Đặt \(\dfrac{1}{cosx}=t\) , do \(x\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow0< cosx< 1\Rightarrow\dfrac{1}{cosx}>1\Rightarrow t>1\)
\(\Rightarrow\left(1-m\right)t^2-2t+4m=0\)
\(\Leftrightarrow\left(2-t\right)\left[\left(m-1\right)t+2m\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2\\\left(m-1\right)t+2m=0\left(1\right)\end{matrix}\right.\)
Bài toán thỏa mãn khi (1) có nghiệm lớn hơn 1
\(\Rightarrow\left\{{}\begin{matrix}m\ne1\\t=\dfrac{2m}{1-m}>1\end{matrix}\right.\) \(\Rightarrow\dfrac{3m-1}{1-m}>0\Rightarrow\dfrac{1}{3}< m< 1\)
