Question

\(cotx-tanx+4sin2x=\frac{2}{sin2x}\)

Answer 1

ĐKXĐ: \(sin2x\ne0\)

\(\Leftrightarrow\frac{cos^2x-sin^2x}{sinx.cosx}+4sin2x=\frac{2}{sin2x}\)

\(\Leftrightarrow\frac{2cos2x}{sin2x}+4sin2x=\frac{2}{sin2x}\)

\(\Leftrightarrow cos2x+2sin^22x=1\)

\(\Leftrightarrow cos2x-2\left(1-cos^22x\right)-1=0\)

\(\Leftrightarrow2cos^22x+cos2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\left(ktm\right)\\cos2x=-\frac{3}{2}>1\left(l\right)\end{matrix}\right.\)

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