1. The problem is to verify if the solutions to a trigonometric equation are $\frac{7\pi}{6}$, $\frac{11\pi}{6}$, and $\frac{\pi}{2}$. 2. Typically, such solutions come from equations involving sine, cosine, or tangent functions where angles are found in the interval $[0, 2\pi)$. 3. Let's assume the equation is $\sin x = -\frac{1}{2}$ or $\cos x = \frac{\sqrt{3}}{2}$ or similar, since these angles correspond to common unit circle values. 4. Check each angle: - $\sin \frac{7\pi}{6} = -\frac{1}{2}$, correct. - $\sin \frac{11\pi}{6} = -\frac{1}{2}$, correct. - $\sin \frac{\pi}{2} = 1$, which is different. 5. If the original equation was $\sin x = -\frac{1}{2}$, then $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ are solutions, but $\frac{\pi}{2}$ is not. 6. If the original equation was $\cos x = \frac{\sqrt{3}}{2}$, then $\frac{\pi}{6}$ and $\frac{11\pi}{6}$ are solutions, not $\frac{7\pi}{6}$ or $\frac{\pi}{2}$. 7. Therefore, the answer depends on the original equation. Without it, we cannot confirm all three angles as solutions. 8. Please provide the original equation for precise verification.