14. Find the exact solutions to \( 2 \sin (2 x)+\sqrt{3}=0 \quad \) for \( x \in[0,2 \pi) \). A) \( \frac{5 \pi}{12} ; \frac{7 \pi}{12} ; \frac{17 \pi}{12} ; \frac{19 \pi}{12} \) B) \( \frac{2 \pi}{3} ; \frac{5 \pi}{6} ; \frac{5 \pi}{3} ; \frac{11 \pi}{6} \) C) \( \frac{\pi}{6} ; \frac{5 \pi}{6} ; \frac{7 \pi}{6} ; \frac{11 \pi}{6} \) D) \( \frac{\pi}{6} ; \frac{\pi}{3} ; \frac{7 \pi}{6} ; \frac{4 \pi}{3} \) E) \( \frac{\pi}{12} ; \frac{5 \pi}{12} ; \frac{13 \pi}{12} ; \frac{17 \pi}{12} \)

To solve the equation \( 2 \sin(2x) + \sqrt{3} = 0 \), we first isolate the sine function: \[ 2 \sin(2x) = -\sqrt{3} \implies \sin(2x) = -\frac{\sqrt{3}}{2}. \] The sine function equals \(-\frac{\sqrt{3}}{2}\) at specific angles: \[ 2x = \frac{4\pi}{3} + 2k\pi \quad \text{and} \quad 2x = \frac{5\pi}{3} + 2k\pi \quad \text{for } k \in \mathbb{Z}. \] Dividing by 2 to solve for \(x\): \[ x = \frac{2\pi}{3} + k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + k\pi. \] For \(k=0\), we find: 1. \(x = \frac{2\pi}{3}\), 2. \(x = \frac{5\pi}{6}\). For \(k=1\) (and keeping within the interval \([0, 2\pi)\)): 3. \(x = \frac{2\pi}{3} + \pi = \frac{5\pi}{3}\), 4. \(x = \frac{5\pi}{6} + \pi = \frac{11\pi}{6}\). Thus, the complete list of solutions is: \[ x = \frac{2\pi}{3}, \frac{5\pi}{6}, \frac{5\pi}{3}, \frac{11\pi}{6}. \] This corresponds to option B) \( \frac{2 \pi}{3} ; \frac{5 \pi}{6} ; \frac{5 \pi}{3} ; \frac{11 \pi}{6} \).