solve the equation for solutions in the interval 0, 2π). sin 4x = √3/2

Explanation:

Step1: Recall the general solution for (\sin\theta=\frac{\sqrt{3}}{2})

We know that if (\sin\theta = \frac{\sqrt{3}}{2}), then (\theta = 2k\pi+\frac{\pi}{3}) or (\theta=(2k + 1)\pi-\frac{\pi}{3}=2k\pi+\frac{2\pi}{3}), where (k\in\mathbb{Z}). Since (\theta = 4x), we have (4x=2k\pi+\frac{\pi}{3}) or (4x=2k\pi+\frac{2\pi}{3}).

Step2: Solve for (x)

For (4x=2k\pi+\frac{\pi}{3}), then (x=\frac{k\pi}{2}+\frac{\pi}{12}). For (4x=2k\pi+\frac{2\pi}{3}), then (x=\frac{k\pi}{2}+\frac{\pi}{6}).

Step3: Find solutions in the interval ([0,2\pi))

When (k = 0):

• For (x=\frac{k\pi}{2}+\frac{\pi}{12}), (x=\frac{\pi}{12})
• For (x=\frac{k\pi}{2}+\frac{\pi}{6}), (x=\frac{\pi}{6}) When (k = 1):
• For (x=\frac{k\pi}{2}+\frac{\pi}{12}), (x=\frac{\pi}{2}+\frac{\pi}{12}=\frac{7\pi}{12})
• For (x=\frac{k\pi}{2}+\frac{\pi}{6}), (x=\frac{\pi}{2}+\frac{\pi}{6}=\frac{2\pi}{3}) When (k = 2):
• For (x=\frac{k\pi}{2}+\frac{\pi}{12}), (x=\pi+\frac{\pi}{12}=\frac{13\pi}{12})
• For (x=\frac{k\pi}{2}+\frac{\pi}{6}), (x=\pi+\frac{\pi}{6}=\frac{7\pi}{6}) When (k = 3):
• For (x=\frac{k\pi}{2}+\frac{\pi}{12}), (x=\frac{3\pi}{2}+\frac{\pi}{12}=\frac{19\pi}{12})
• For (x=\frac{k\pi}{2}+\frac{\pi}{6}), (x=\frac{3\pi}{2}+\frac{\pi}{6}=\frac{5\pi}{3})

Answer:

B. (\left{\frac{\pi}{12},\frac{\pi}{6},\frac{2\pi}{3},\frac{7\pi}{12},\frac{7\pi}{6},\frac{13\pi}{12},\frac{5\pi}{3},\frac{19\pi}{12}\right})