Given:
[tex]\begin{gathered} 4sin^2\theta-4=-3 \\ 0\leq\theta<2\pi \end{gathered}[/tex]To determine the value(s) for θ, we follow first let sin θ equal to u. Hence,
[tex]\begin{gathered} 4sin^2\theta-4=-3 \\ 4u^2-4=-3 \\ Simplify\text{ and rearrange} \\ 4u^2=-3+4 \\ 4u^2=1 \\ u^2=\frac{1}{4} \\ u=\pm\sqrt{\frac{1}{4}} \\ Calculate \\ u=\frac{1}{2},\text{ u}=-\frac{1}{2} \end{gathered}[/tex]We substitute back u= sin θ. So,
[tex]sin\theta=\frac{1}{2},sin\theta=-\frac{1}{2}[/tex]Now, we consider the range:
[tex]\begin{gathered} For\text{ }sin\theta=\frac{1}{2},0\leq\theta<2\pi: \\ \theta=\frac{\pi}{6},\theta=\frac{5\pi}{6} \end{gathered}[/tex][tex]\begin{gathered} For\text{ s}\imaginaryI n\theta=-\frac{1}{2},0\leq\theta\lt2\pi: \\ \theta=\frac{7\pi}{6},\theta=\frac{11\pi}{6} \end{gathered}[/tex]Therefore, the answers are:
[tex]\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}[/tex]