The principal solutions of sqrt{3} sec x + 2 | vedclass

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(B) Given equation: $\sqrt{3} \sec x + 2 = 0$ $\sec x = -\frac{2}{\sqrt{3}}$ $\cos x = -\frac{\sqrt{3}}{2}$ Since $\cos x$ is negative in the second a

Vedclass · Accepted Answer

$\frac{5 \pi}{6}, \frac{7 \pi}{6}$

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(B) Given equation: $\sqrt{3} \sec x + 2 = 0$ $\sec x = -\frac{2}{\sqrt{3}}$ $\cos x = -\frac{\sqrt{3}}{2}$ Since $\cos x$ is negative in the second and third quadrants,we find the principal solutions in the interval $[0, 2\pi)$. $\cos x = -\cos(\frac{\pi}{6}) = \cos(\pi - \frac{\pi}{6}) = \cos(\frac{5\pi}{6})$ $\cos x = \cos(\pi + \frac{\pi}{6}) = \cos(\frac{7\pi}{6})$ Thus,the principal solutions are $x = \frac{5\pi}{6}$ and $x = \frac{7\pi}{6}$.

The principal solutions of $\sqrt{3} \sec x + 2 = 0$ are

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