Sinx + cosx-sinxcosx=1

Đáp án:

$\left[\begin{array}{l}x =k2\pi\\x = \dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

Giải thích các bước giải:

$\sin x + \cos x -\sin x\cos x = 1$

$\Leftrightarrow\sin x(1-\cos x) + \cos x -1=0$

$\Leftrightarrow (\cos x -1)(1-\sin x) = 0$

$\Leftrightarrow \left[\begin{array}{l}\cos x = 1\\\sin x = 1\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{l}x =k2\pi\\x = \dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$