Solve the equation on the interval \( 0 \leq \theta<2 \pi \) \( 4 \sin ^{2} \theta=1 \)

To solve the equation \( 4 \sin^{2} \theta = 1 \), we start by rewriting it in a simpler form: \[ \sin^{2} \theta = \frac{1}{4} \] Now, taking the square root of both sides gives us: \[ \sin \theta = \frac{1}{2} \quad \text{or} \quad \sin \theta = -\frac{1}{2} \] Next, we look for the angles that satisfy these sine values within the interval \( 0 \leq \theta < 2 \pi \). 1. For \( \sin \theta = \frac{1}{2} \): - This occurs at: \[ \theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{5\pi}{6} \] 2. For \( \sin \theta = -\frac{1}{2} \): - This occurs at: \[ \theta = \frac{7\pi}{6} \quad \text{and} \quad \theta = \frac{11\pi}{6} \] Combining all solutions, we find: \[ \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \] So, the complete set of solutions for the equation \( 4 \sin^{2} \theta = 1 \) in the interval \( 0 \leq \theta < 2\pi \) is: \[ \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \]