Giải mình với mn ơi....

Giải thích các bước giải:

$\left ( 1 - \tan{x} \right )\left ( 1 + \sin{2x} \right ) = 1 + \tan{x}$ (ĐK $\cos{x} \neq 0$)

$\Leftrightarrow \left ( 1 - \dfrac{\sin{x}}{\cos{x}} \right )\left ( \sin^{2}{x} + 2\sin{x}\cos{x} + \cos^{2}{x} \right ) = 1 + \dfrac{\sin{x}}{\cos{x}}$

$\Leftrightarrow \dfrac{\left ( \cos^{2}x - \sin^{2}x \right )\left ( \sin{x} + \cos{x} \right )}{\cos{x}} = \dfrac{\sin{x} + \cos{x}}{\cos{x}}$

$\Leftrightarrow \cos{2x}\left ( \sin{x} + \cos{x} \right ) = \sin{x} + \cos{x}$

$\Leftrightarrow \sqrt{2}{\left ( x + \dfrac{\pi}{4} \right )}\sin\left ( \cos{2x} - 1 \right ) = 0$

$\Leftrightarrow \left[ \begin{array}{l}\sin{\left ( x + \dfrac{\pi}{4} \right )} = 0\\\cos{2x} = 1\end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l}x + \dfrac{\pi}{4} = k\pi\\2x = k2\pi\end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l}x  = -\dfrac{\pi}{4} + k \pi (tm)\\x = k\pi (tm)\end{array} \right.$