Đáp án:
$6) x=0\\ 7)x=\dfrac{\pi}{2}.$
Giải thích các bước giải:
$6)\\ \sin^2x+\sin x=0\\ \Leftrightarrow \sin x(\sin x+1)=0\\ \Leftrightarrow \left[\begin{array}{l} \sin x=0 \\ \sin x=-1\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=k \pi(k \in \mathbb{Z}) \\ x=-\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z}) \end{array} \right.\\ -\dfrac{\pi}{2} < x < \dfrac{\pi}{2} \Rightarrow x=0\\ 7)\\ \cos^2x-\cos x=0\\ \Leftrightarrow \cos x(\cos x-1)=0\\ \Leftrightarrow \left[\begin{array}{l} \cos x=0 \\ \cos x=1\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=\dfrac{\pi}{2}+k \pi(k \in \mathbb{Z}) \\ x=k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\ 0 < x < \pi \Rightarrow x=\dfrac{\pi}{2}.$
