Solve   \( 2sin^2x+3cosx-3=0 \)   in the interval \( \quad 0^{\circ} \leqslant x \leqslant 180^{\circ}\)

\begin{align} 0 & = 2sin^2x+3cos-3 \\ 0 & = 2(1-cos^2x)+3cosx-3 \\ 0 & = -2cos^2x +2 +3cosx-3\\ 0 & = 2cos^2x-3cosx+1 \\ 0 & = 2cos^2x -2cosx-cosx+1 \\ 0 & =2cosx(cosx-1)-(cosx-1) \\ 0 & = (2cosx-1)(cosx-1)\\ \implies cosx & = \frac{1}{2} \quad and \quad cosx=1 \\ \therefore\ \quad x & = 0^{\circ},\ and \quad x= 60^{\circ} \\ \\ \end{align}