solve for exact solutions over the interval $0,2\\pi)$.\n$\\sin 2x = - \\frac{\\sqrt{3}}{2}$\nselect the…

Explanation:

Step1: Find the general solutions for (2x)

We know that (\sin\theta=-\frac{\sqrt{3}}{2}). The solutions for (\theta) (where (\theta = 2x)) in the unit - circle are (\theta=\frac{4\pi}{3}+2k\pi) and (\theta=\frac{5\pi}{3}+2k\pi), (k\in\mathbb{Z}) (since (\sin\theta = y) - coordinate on the unit circle and (\sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2},\sin\frac{5\pi}{3}=-\frac{\sqrt{3}}{2})).

So, (2x=\frac{4\pi}{3}+2k\pi) or (2x=\frac{5\pi}{3}+2k\pi), (k\in\mathbb{Z}).

Step2: Solve for (x)

For (2x=\frac{4\pi}{3}+2k\pi), divide both sides by (2): (x=\frac{2\pi}{3}+k\pi)

For (2x=\frac{5\pi}{3}+2k\pi), divide both sides by (2): (x=\frac{5\pi}{6}+k\pi)

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

When (k = 0): For (x=\frac{2\pi}{3}+k\pi), (x=\frac{2\pi}{3}) For (x=\frac{5\pi}{6}+k\pi), (x=\frac{5\pi}{6})

When (k = 1): For (x=\frac{2\pi}{3}+k\pi), (x=\frac{2\pi}{3}+\pi=\frac{5\pi}{3}) For (x=\frac{5\pi}{6}+k\pi), (x=\frac{5\pi}{6}+\pi=\frac{11\pi}{6})

Answer:

The solution set is (\frac{5\pi}{6},\frac{2\pi}{3},\frac{11\pi}{6},\frac{5\pi}{3})