ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)
Pt tương đương:
\(\dfrac{sinx-cosx}{sinx.cosx}=2\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\dfrac{-\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)}{sinx.cosx}=2\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x+\dfrac{\pi}{4}\right)=0\\2sinx.cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x+\dfrac{\pi}{4}\right)=0\\sin2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\\2x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{4}+k\pi\)
