91Ó°ÊÓ

To remove the square, we will take the square root of both sides of the equation. Remember that when we take the square root of a squared expression, we need to take both the positive and negative square roots. Therefore, we will have two separate equations: $$\sin\theta=\pm\sqrt{\frac{1}{4}}$$ This simplifies to: $$\sin\theta=\pm\frac{1}{2}$$

Now, we need to find all the angles \(\theta\) in the range of \(0\leq\theta<2\pi\) that satisfy \(\sin\theta=\frac{1}{2}\) or \(\sin\theta=-\frac{1}{2}\). For \(\sin\theta=\frac{1}{2}\), we know that \(\theta=\frac{\pi}{6}\) and \(\theta=\frac{5\pi}{6}\), since these are the two angles in the given range for which the sine function equals \(\frac{1}{2}\). For \(\sin\theta=-\frac{1}{2}\), we know that \(\theta=\frac{7\pi}{6}\) and \(\theta=\frac{11\pi}{6}\), since these are the two angles in the given range for which the sine function equals \(-\frac{1}{2}\). Therefore, the solutions of the equation \(\sin^{2}\theta=\frac{1}{4}\) in the range \(0\leq\theta<2\pi\) are: $$\theta=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$$