1) The solution of sin x = $-\frac{\sqrt{3}}{2}$ is A) $x = n\pi+ (-1)^{n}\frac{4\pi}{3}, n \epsilon Z$ B) x = n\pi+ (-1)^{n}\frac{2\pi}{3}, n \epsilon Z C) $x = n\pi+ (-1)^{n}\frac{3\pi}{3}, n \epsilon Z$ D) None of these Answer: Option AExplanation:We have, sin x = $-\frac{\sqrt{3}}{2}$ = -sin$\frac{\pi}{3}$ Hence, sin x = $\frac{4\pi}{3}$ which gives $x = n\pi+ (-1)^{n}\frac{4\pi}{3}, n \epsilon Z$
