Giải phương trình $cotx - 1 = \frac{\cos 2x}{1 + \tan x} + {\sin }^{2}x - \frac{1}{2}\sin 2x$

Điều kiện: \[\left\{ \begin{array}{l}\sin x \ne 0\\\cos x \ne 0\\\tan x \ne  - 1\end{array} \right.\]

\[\cot x - 1 = \frac{{\cos 2x}}{{1 + \tan x}} + {\sin ^2}x - \frac{1}{2}\sin 2x\]

\[ \Leftrightarrow \frac{{\cos x}}{{\sin x}} - 1 = \frac{{\cos 2x}}{{1 + \frac{{sinx}}{{\cos x}}}} + {\sin ^2}x - \frac{1}{2}\sin 2x\]

\[ \Leftrightarrow \frac{{\cos x - \sin x}}{{\sin x}} = \frac{{\cos 2x\,.\,\,\cos x\,}}{{\cos x + sinx}} + {\sin ^2}x - \sin x\,\cos x\]

\[ \Leftrightarrow \frac{{\cos x - \sin x}}{{\sin x}} = \frac{{({{\cos }^2}x - si{n^2}x)\,.\,\,\cos x\,}}{{\cos x + sinx}} + \sin x(\sin x - \,\cos x)\]

\[ \Leftrightarrow \frac{{\cos x - \sin x}}{{\sin x}} = \frac{{(\cos x + sinx)(\cos x - sinx)\,.\,\,\cos x\,}}{{\cos x + sinx}} + \sin x(\sin x - \,\cos x)\]

\[ \Leftrightarrow \frac{{\cos x - \sin x}}{{\sin x}} = (\cos x - sinx)\,.\,\,\cos x + \sin x(\sin x - \,\cos x)\]

\[ \Leftrightarrow \frac{{\cos x - \sin x}}{{\sin x}} - (\cos x - sinx)\,.\,\,\cos x + \sin x(\cos x - sinx) = 0\]

\[ \Leftrightarrow (\cos x - sinx)\left[ {\frac{1}{{\sin x}} - \,\cos x + \sin x} \right] = 0\]

\[ \Leftrightarrow \left[ \begin{array}{l}\cos x - sinx = 0\\\frac{1}{{\sin x}} - \,\cos x + \sin x = 0\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}\cos x = sinx\\1 - \,\cos x\,.\,\sin x + {\sin ^2}x = 0\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}\cos x = sinx\\1 - \,\frac{1}{2}\sin 2x + \frac{{1 - \cos 2x}}{2} = 0\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}\tan x = 1\\\sin 2x + \cos 2x = 3\,\,(VL)\end{array} \right.\]

\[ \Leftrightarrow x = \frac{\pi }{4} + k\pi \,\,(k \in \mathbb{Z})\]

Kết hợp với điều kiện ta suy ra phương trình có nghiệm \[x = \frac{\pi }{4} + k\pi \,\,(k \in \mathbb{Z})\].