Solve   \( 2sin^2x-5sinx-3=0 \)   in the interval \( \quad -90^{\circ} \leqslant x \leqslant 270^{\circ}\)

\begin{align} 0 & = 2sin^2x-5sinx-3 \\ 0 & = 2sin^2x-6sinx+sinx-3 \\ 0 & = 2sinx(sinx-3)+(sinx-3) \\ 0 & = (2sinx+1)(sinx-3) \\ \implies sinx= & -\frac{1}{2} \\ or \quad sinx & =3\ (impossible,\ so\ reject\ this\ solution) \\ \\ \therefore\ \quad x & = -30^{\circ},\ and \quad x= 210^{\circ} \\ \\ \end{align}