Question

giair pt:\(\frac{1-\cos x\cos2x}{\sin2x}-\frac{1}{\cos x}=4\sin^2x-\sin x-1\)

Answer 1

ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(\Leftrightarrow\frac{1-cosx.cos2x-2sinx}{sin2x}=1-sinx-2cos2x\)

\(\Leftrightarrow1-cosx.cos2x-2sinx=sin2x-sinx.sin2x-2cos2x.sin2x\)

\(\Leftrightarrow1-2sinx=sin2x+\left(cos2x.cosx-sin2x.sinx\right)-sin4x\)

\(\Leftrightarrow1-2sinx=sin2x+cos3x-sin4x\)

\(\Leftrightarrow1-2sinx=cos3x-2cos3x.sinx\)

\(\Leftrightarrow1-2sinx=cos3x\left(1-2sinx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=1\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)