Đk: \(x\ne k\pi\) (\(k\in\)\(Z\))
\(Pt\Leftrightarrow\dfrac{\left(sinx^2+cos^2x\right)^2-2sin^2x.cos^2x}{sinx}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1-\dfrac{1}{2}sin^22x}{sinx}=\dfrac{1}{4}\)
\(\Leftrightarrow4-2sin^22x=sinx\)
\(\Leftrightarrow4-2\left(1-cos^22x\right)=sinx\)
\(\Leftrightarrow2+2\left[1-\left(1-2sin^2x\right)^2\right]=sinx\)
\(\Leftrightarrow-8sin^4x+8sin^2x-sinx+2=0\)
\(\Leftrightarrow-8sin^2x\left(sin^2x-1\right)-sinx+2=0\)
Có \(\left\{{}\begin{matrix}0< sin^2x\le1\\sinx\le1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}sin^2x-1\le0\\-sinx\ge-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-8sin^2x\left(sin^2x-1\right)\ge0\\2-sinx\ge1\end{matrix}\right.\)
\(\Rightarrow-8sin^2x\left(sin^2x-1\right)-sinx+2\ge1\)>0
=> Dấu bằng không xảy ra,
Vậy pt vô nghiệm.
