$\begin{align} & \text{Obtain the first general solution of the equation and then use the general } \\ & \text{solution to find all solution in the range }{{0}^{\circ }}\le \theta \le {{360}^{\circ }} \\ & 8{{\sin }^{2}}\theta -6\cos \theta =3 \\\end{align}$
$\begin{align} & 8{{\sin }^{2}}\theta -6\cos \theta =3 \\ & 8(1-{{\cos }^{2}}\theta )-6\cos \theta -3=0 \\ & 8-8{{\cos }^{2}}\theta -6\cos \theta -3=0 \\ & 8{{\cos }^{2}}\theta +6\cos \theta -5=0 \\ & (2\cos \theta -1)(4\cos \theta +5)=0 \\ & \cos \theta =\frac{1}{2}\text{ or }\cos \theta =-\frac{5}{4} \\ & \text{Reject }\cos \theta =-\frac{5}{4}\text{ because }-1\le \cos \theta \le 1 \\ & \theta ={{\cos }^{-1}}\frac{1}{2}={{60}^{\circ }} \\ & \text{The general solution }\theta =n{{360}^{\circ }}\pm {{60}^{\circ }} \\ & \theta ={{60}^{\circ }},{{300}^{\circ }} \\\end{align}$