PTLG đưa về PT tích.

Đề bài (Câu hỏi của bạn Minh Phương Ngô hỏi trên facebook Trợ Giúp Toán Học)
Giải phương trình: $\ \frac{{2\left[ {{{\left( {sin\,x - cos\,x} \right)}^2} + 2si{n^2}x} \right]}}{{1 + ta{n^2}x}} + sin\left( {x + \frac{\pi }{4}} \right) + cos\left( {3x - \frac{\pi }{4}} \right) = 2 + 2\,cos2x\left( * \right).$
Giải: Điều kiện $\ x \ne \frac{\pi }{2} + k\pi ,k \in Z.$
$$\begin{array}{l} \left( * \right) \Leftrightarrow 2co{s^2}x\left[ {{{\left( {sin\,x - cos\,x} \right)}^2} + 2si{n^2}x} \right] + cos\left( {\frac{\pi }{4} - x} \right) + cos\left( {3x - \frac{\pi }{4}} \right) = 2 + 2\,cos2x\\  \Leftrightarrow \left( {cos2x + 1} \right)\left( {1 - sin\,2x + 1 - cos\,2x} \right) + 2cos\,xcos\,\left( {2x - \frac{\pi }{4}} \right) = 2\left( {cos2x + 1} \right)\\  \Leftrightarrow \left( {cos2x + 1} \right)\left[ {2 - \left( {cos\,2x + sin\,2x} \right)} \right] + 2cos\,xcos\,\left( {2x - \frac{\pi }{4}} \right) = 2\left( {cos2x + 1} \right)\\  \Leftrightarrow 2\left( {cos2x + 1} \right) - \sqrt 2 \left( {cos2x + 1} \right)cos\,\left( {2x - \frac{\pi }{4}} \right) + 2cos\,xcos\,\left( {2x - \frac{\pi }{4}} \right) = 2\left( {cos2x + 1} \right)\\  \Leftrightarrow \sqrt 2 cos\,\left( {2x - \frac{\pi }{4}} \right)\left( {\sqrt 2 \cos x\, - cos2x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} cos\,\left( {2x - \frac{\pi }{4}} \right) = 0\\ \sqrt 2 \cos x\left( {1 - \sqrt 2 \cos x} \right) = 0 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} cos\,\left( {2x - \frac{\pi }{4}} \right) = 0 \Leftrightarrow 2x - \frac{\pi }{4} = \frac{\pi }{2} + k\pi  \Leftrightarrow x = \frac{{3\pi }}{8} + \frac{{k\pi }}{2}\\ cos\,x = \frac{1}{{\sqrt 2 }} \Leftrightarrow x =  \pm \frac{\pi }{4} + k2\pi \end{array} \right. \Rightarrow S = \left\{ {\frac{{3\pi }}{8} + \frac{{k\pi }}{2}; \pm \frac{\pi }{4} + k2\pi ,k \in Z} \right\} \end{array}$$