The principal solutions of the equation sqrt{ | vedclass

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Question

(C) Given the equation $\sqrt{3} \sec x + 2 = 0$. $\sqrt{3} \sec x = -2$ $\sec x = -\frac{2}{\sqrt{3}}$ $\cos x = -\frac{\sqrt{3}}{2}$ Since $\cos x$

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

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

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

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