Question

Giải phương trình:

\(sin^2x.tanx+cos^2xcotx-sin2x=1+tanx+cots\)

Answer 1

ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\frac{k\pi}{2}\)

\(\Leftrightarrow tanx-sin^2x.tanx+cotx-cos^2x.cotx+1+sin2x=0\)

\(\Leftrightarrow tanx\left(1-sin^2x\right)+cotx\left(1-cos^2x\right)+1+sin2x=0\)

\(\Leftrightarrow\frac{sinx}{cosx}.cos^2x+\frac{cosx}{sinx}.sin^2x+1+sin2x=0\)

\(\Leftrightarrow2sinx.cosx+1+sin2x=0\)

\(\Leftrightarrow2sin2x=-1\Rightarrow sin2x=-\frac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x=-\frac{\pi}{3}+k2\pi\\2x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-\pi}{6}+k\pi\\x=\frac{2\pi}{3}+k\pi\end{matrix}\right.\)