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) \( \frac{\pi}{12} ; \frac{5 \pi}{12} ; \frac{13 \pi}{12} ; \frac{17 \pi}{12} \)

We begin with the equation:   2 sin(2x) + √3 = 0 Step 1: Isolate sin(2x).   2 sin(2x) = -√3   sin(2x) = -√3/2 Step 2: Find the angles for which sin θ = -√3/2.  sin θ = -√3/2 when θ = 4π/3 or θ = 5π/3 (in the interval [0, 2π)). Since the sine function is periodic with period 2π, the general solution is:   θ = 4π/3 + 2πk or θ = 5π/3 + 2πk (where k is any integer). Step 3: Replace θ with 2x.   2x = 4π/3 + 2πk or 2x = 5π/3 + 2πk Step 4: Solve for x by dividing both sides by 2.  For the first case:   x = (4π/3) ÷ 2 + πk = 2π/3 + πk  For the second case:   x = (5π/3) ÷ 2 + πk = 5π/6 + πk Step 5: Find the solutions for x in [0, 2π). For x = 2π/3 + πk:  - When k = 0: x = 2π/3  - When k = 1: x = 2π/3 + π = 5π/3  - k = -1 gives negative values, so we discard. For x = 5π/6 + πk:  - When k = 0: x = 5π/6  - When k = 1: x = 5π/6 + π = 11π/6  - k = -1 gives a negative value, so we discard. Thus, the four solutions within [0, 2π) are:  x = 2π/3, 5π/6, 5π/3, 11π/6 Matching these with the provided options, we see that they correspond exactly to option B. Answer: Option B.