Mathematics | VITEEE 2019 | Discussion Page

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The solution of sin x = $-\frac{\sqrt{3}}{2}$ is

$x = n\pi+ (-1)^{n}\frac{4\pi}{3}, n \epsilon Z$

B) x = n\pi+ (-1)^{n}\frac{2\pi}{3}, n \epsilon Z

$x = n\pi+ (-1)^{n}\frac{3\pi}{3}, n \epsilon Z$

D) None of these

Option A

We have, sin x = $-\frac{\sqrt{3}}{2}$

= -sin $\frac{\pi}{3}$

Hence, sin x = $\frac{4\pi}{3}$ which gives

$x = n\pi+ (-1)^{n}\frac{4\pi}{3}, n \epsilon Z$

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