cot x - 1 = cos2x/(1+tanx) +sin x^{2} -1/2 . sin2x mn giải giúp nha cam on nhieu

Đáp án:

$x = \dfrac{\pi }{4} + k\pi $ $(k\in\mathbb Z)$

Lời giải:

$\begin{array}{l} \cot x - 1 = \dfrac{{\cos 2x}}{{1 + \tan x}} + {\sin ^2}x - \dfrac{1}{2}\sin 2x\\ \Leftrightarrow \dfrac{{\cos x}}{{\sin x}} - 1 = \dfrac{{\cos 2x}}{{1 + \dfrac{{\sin x}}{{\cos x}}}} + {\sin ^2}x - \dfrac{1}{2}\sin 2x\\ \Leftrightarrow \dfrac{{\cos x}}{{\sin x}} - 1 = \left( {\cos x - \sin x} \right)\cos x + {\sin ^2}x - \sin x\cos x\\ \Leftrightarrow \dfrac{{\cos x - \sin x}}{{\sin x}} = \left( {\cos x - \sin x} \right)\cos x + \sin \left( {\cos x - \sin x} \right)\\ \Leftrightarrow \left( {\cos x - \sin x} \right)\left[ {\dfrac{1}{{\sin x}} - \cos x + \sin x} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos x - \sin x = 0\\ 1 - \cos x\sin x + {\sin ^2}x = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \cos x = \sin x\\ 1 - \dfrac{1}{2}\sin 2x + \dfrac{{1 - \cos 2x}}{2} = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \tan x = 1\\ 2 - \sin 2x + 1 - \cos 2x = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \tan x = 1\\ \sin 2x + \cos 2x = 3\text{ (vô nghiệm)}\end{array} \right. \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \end{array}$ $(k\in\mathbb Z)$