Question

\(cotx-1=\frac{cos2x}{1+tanx}+sin^2x-\frac{1}{2}sin2x\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}sin2x\ne0\\tanx\ne-1\end{matrix}\right.\)

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

\(\Leftrightarrow\frac{cosx-sinx}{sinx}=cosx\left(cosx-sinx\right)-sinx\left(cosx-sinx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Rightarrow x=\frac{\pi}{4}+k\pi\\\frac{1}{sinx}=cosx-sinx\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow sinx.cosx-sin^2x=1\)

\(\Leftrightarrow2sinx.cosx+1-2sin^2x=3\)

\(\Leftrightarrow sin2x+cos2x=3\)

Vế trái không lớn hơn 2 nên pt vô nghiệm