Cho phương trình 2cos² x + sin x -1 =0 có bao nhiêi nghiệm trên khoảng (0;6)

$\begin{array}{l} 2{\cos ^2}x + \sin x - 1 = 0\\ \Leftrightarrow 2\left( {1 - {{\sin }^2}x} \right) + \sin x - 1 = 0\\ \Leftrightarrow - 2{\sin ^2}x + \sin x + 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = 1\\ \sin x = - \frac{1}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = - \frac{\pi }{6} + k2\pi \\ x = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\\ Cho\,0 < \frac{\pi }{2} + k2\pi < 6\,tim\,k.\\ Cho\,0 < - \frac{\pi }{6} + k2\pi < 6\,tim\,k.\\ Cho\,0 < \frac{{7\pi }}{6} + k2\pi < 6\,tim\,k.\\ Tu\,do\,ket\,luan\,ban\,nhe! \end{array}$