Đáp án:
$S=\{\ \dfrac{\pi}{2}+k\pi ;-\dfrac{\pi}{2}+k2\pi |k\in Z \}$
Giải thích các bước giải:
$sinx.cosx+cosx=0$
$cosx(sinx+1)=0$
\(\left[ \begin{array}{l}cosx=0\\sinx+1=0\end{array} \right.\)
\(\left[ \begin{array}{l}cosx=0\\sinx=-1\end{array} \right.\)
\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{array} \right.,k\in Z\)
Vậy $S=\{\ \dfrac{\pi}{2}+k\pi ;-\dfrac{\pi}{2}+k2\pi |k\in Z \}$
