Solve the equation on the interval $0 \leq \theta< 2 \pi$.
\[
4 \sin ^{2} \theta-1=0
\]

Step 1 ：The given equation is a quadratic equation in terms of \(\sin \theta\). We can solve this equation by setting it equal to zero and solving for \(\sin \theta\). Once we have the values of \(\sin \theta\), we can find the corresponding values of \(\theta\) in the given interval.

Step 2 ：The given equation is \(4\sin^2\theta - 1 = 0\). We can rewrite this as \(\sin^2\theta = \frac{1}{4}\), which gives \(\sin\theta = \pm\frac{1}{2}\).

Step 3 ：The solutions to this equation in the interval \(0 \leq \theta < 2\pi\) are \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\).

Step 4 ：Final Answer: \(\boxed{\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}}\)