Find the principal solutions of the equation | vedclass

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Question

(A) We know that $	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}.$ Since $	an x$ is negative in the second and fourth quadrants,we have: $	an x = 	an(\pi

Vedclass · Accepted Answer

$\frac{5 \pi}{6}$ and $\frac{11 \pi}{6}$

Vedclass · Answer

(A) We know that $	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}.$ Since $	an x$ is negative in the second and fourth quadrants,we have: $	an x = 	an(\pi - \frac{\pi}{6}) = 	an \frac{5 \pi}{6} = -\frac{1}{\sqrt{3}}$ and $	an x = 	an(2 \pi - \frac{\pi}{6}) = 	an \frac{11 \pi}{6} = -\frac{1}{\sqrt{3}}.$ Thus,the principal solutions are $\frac{5 \pi}{6}$ and $\frac{11 \pi}{6}.$

Find the principal solutions of the equation $\tan x = -\frac{1}{\sqrt{3}}.$

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