ĐKXĐ: ...
\(\Leftrightarrow48-\frac{1}{cos^4x}-\frac{2}{sin^2x}\left(1+\frac{cosx.cos2x}{sinx.sin2x}\right)=0\)
\(\Leftrightarrow48-\frac{1}{cos^4x}-\frac{2}{sin^2x}\left(\frac{cos2x.cosx+sin2x.sinx}{2sin^2x.cosx}\right)=0\)
\(\Leftrightarrow48-\frac{1}{cos^4x}-\frac{2}{sin^2x}.\frac{cosx}{2sin^2x.cosx}=0\)
\(\Leftrightarrow48-\frac{1}{cos^4x}-\frac{1}{sin^4x}=0\)
\(\Leftrightarrow\frac{sin^4x+cos^4x}{sin^4x.cos^4x}=48\Leftrightarrow\frac{1-2sin^2x.cos^2x}{sin^4x.cos^4x}=48\)
\(\Leftrightarrow\frac{1-\frac{1}{2}sin^22x}{\frac{1}{16}sin^42x}=48\)
Đặt \(sin^22x=t\Rightarrow0< t\le1\)
\(\Rightarrow\frac{1-\frac{1}{2}t}{\frac{1}{16}t^2}=48\Leftrightarrow3t^2+\frac{1}{2}t-1=0\)
\(\Rightarrow t=\frac{1}{2}\Rightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos4x=\frac{1}{2}\Leftrightarrow cos4x=0\)
\(\Leftrightarrow...\)
