\(2cos^2x-5sinx-2=0\)
\(\Rightarrow2\left(1-sin^2x\right)-5sinx-2=0\)
\(\Rightarrow-2sin^2x-5sinx=0\)
\(\Rightarrow\left\{{}\begin{matrix}sinx=0\\2sinx+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=k\pi\\sinx=-\dfrac{5}{2}\left(l\right)\end{matrix}\right.\)\(\left(k\in Z\right)\)