19)
Pt \(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}\left[-1+sin\left(2x-\pi\right)\right]=\dfrac{1}{2}\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[-1-sin2x\right]=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{-1}{2}sin^22x-\dfrac{1}{2}sin2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=k\dfrac{\pi}{2}\end{matrix}\right.\),\(k\in Z\)
Vậy...