If theta in left[-frac{7 pi}{6}, frac{4 pi}{3 | vedclass

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(A) Let $x = \operatorname{cosec} 	heta$. The equation becomes $\sqrt{3}x^2 - 2(\sqrt{3}-1)x - 4 = 0$. Using the quadratic formula $x = \frac{-b \pm

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$6$

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(A) Let $x = \operatorname{cosec} 	heta$. The equation becomes $\sqrt{3}x^2 - 2(\sqrt{3}-1)x - 4 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{2(\sqrt{3}-1) \pm \sqrt{4(\sqrt{3}-1)^2 + 16\sqrt{3}}}{2\sqrt{3}}$ $x = \frac{2(\sqrt{3}-1) \pm \sqrt{4(3+1-2\sqrt{3}) + 16\sqrt{3}}}{2\sqrt{3}} = \frac{2(\sqrt{3}-1) \pm \sqrt{16+8\sqrt{3}}}{2\sqrt{3}}$. Since $\sqrt{16+8\sqrt{3}} = \sqrt{(2\sqrt{3}+2)^2} = 2\sqrt{3}+2$,we get: $x = \frac{2\sqrt{3}-2 \pm (2\sqrt{3}+2)}{2\sqrt{3}}$. Case $1$: $x = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 \implies \sin 	heta = \frac{1}{2}$. Case $2$: $x = \frac{-4}{2\sqrt{3}} = -\frac{2}{\sqrt{3}} \implies \sin 	heta = -\frac{\sqrt{3}}{2}$. In the interval $	heta \in [-\frac{7\pi}{6}, \frac{4\pi}{3}]$,$\sin 	heta = \frac{1}{2}$ has $3$ solutions: $\frac{\pi}{6}, \frac{5\pi}{6}, -\frac{7\pi}{6}$. $\sin 	heta = -\frac{\sqrt{3}}{2}$ has $3$ solutions: $-\frac{\pi}{3}, \frac{4\pi}{3}, -\frac{2\pi}{3}$. Total number of solutions = $3 + 3 = 6$.

If $\theta \in \left[-\frac{7 \pi}{6}, \frac{4 \pi}{3}\right]$,then the number of solutions of $\sqrt{3} \operatorname{cosec}^2 \theta - 2(\sqrt{3}-1) \operatorname{cosec} \theta - 4 = 0$ is equal to

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