Question

[Yuri]

\(\dfrac{1}{cosx\left(sinx-cosx\right)}=2\sqrt{2}sinx+\dfrac{2cosx}{sinx-cosx}\)

Bài này giải như nào ạ ??? Em cảm ơn trước nha.

Answer 1

[Minhmetmoi]

Đk: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+m2\pi\\x\ne\dfrac{\pi}{4}+n\pi\end{matrix}\right.\left(m,n\in Z\right)\)

PT \(\Leftrightarrow1=2\sqrt{2}sinx.cosx\left(sinx-cosx\right)+2cos^2x\)

\(\Leftrightarrow\sqrt{2}.2sinx.cosx\left(sinx-cosx\right)+\left(2cos^2x-1\right)=0\)

\(\Leftrightarrow\sqrt{2}sin2x\left(sinx-cosx\right)+\left(cosx-sinx\right)\left(cosx+sinx\right)=0\)

\(\Leftrightarrow\sqrt{2}sin2x=sinx+cosx\)

\(\Leftrightarrow\sqrt{2}sin2x=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\dfrac{\pi}{4}+k2\pi\\2x=\pi-x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{4}+k\dfrac{2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)

Answer 2

[Yuri]

Cảm mơn nhiều nha :3