The principal solutions of the equation sqrt{ | vedclass

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(B) Given the equation $\sqrt{3} \operatorname{cosec} x + 2 = 0$. Rearranging the terms,we get $\operatorname{cosec} x = -\frac{2}{\sqrt{3}}$. Since $

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$\frac{4 \pi}{3}, \frac{5 \pi}{3}$

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(B) Given the equation $\sqrt{3} \operatorname{cosec} x + 2 = 0$. Rearranging the terms,we get $\operatorname{cosec} x = -\frac{2}{\sqrt{3}}$. Since $\operatorname{cosec} x = \frac{1}{\sin x}$,we have $\sin x = -\frac{\sqrt{3}}{2}$. We know that $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$. Since $\sin x$ is negative in the third and fourth quadrants,the principal solutions are: $x = \pi + \frac{\pi}{3} = \frac{4 \pi}{3}$ $x = 2 \pi - \frac{\pi}{3} = \frac{5 \pi}{3}$ Thus,the principal solutions are $\frac{4 \pi}{3}$ and $\frac{5 \pi}{3}$.

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

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