ĐK: $sinx+cosx \neq 0$
pt $\Leftrightarrow \left ( 1-sin^2x \right ).\left ( cosx-1 \right )=2\left ( 1+sinx \right ).\left ( sinx+cosx \right ) $
$\Leftrightarrow \left[ \begin{array}{l} 1+sinx=0\\ \left ( 1-sinx \right ).\left ( cosx-1 \right )=2.\left ( sinx+cosx \right ) \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} sinx=-1\\ cosx+sinx-cosx.sinx-1 = 2.\left ( sinx+cosx \right ) \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{\pi}{2}+k2 \pi\\ \left(cosx+sinx \right) + cosx.sinx + 1 = 0 \end{array} \right.$
Đặt $t=(sinx+cosx) \neq 0$ , khi đó $\dfrac{t^2 -1}{2} = sinx.cosx$
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