If [-(7 )/(6), (4 )/(3)], then the number of solutions of √3 cosec^2 -2(√3-1) cosec -4=0, …

Question

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

Accepted Answer

To find the solutions of the equation $\sqrt{3} \operatorname{cosec}^2 	heta - 2(\sqrt{3} - 1) \operatorname{cosec} 	heta - 4 = 0$, we can begin by solving for $\operatorname{cosec} 	heta$.

The quadratic equation in terms of $\operatorname{cosec} 	heta$ is:

$ \sqrt{3} \operatorname{cosec}^2 	heta - 2(\sqrt{3} - 1) \operatorname{cosec} 	heta - 4 = 0 $

Using the quadratic formula:

$ \operatorname{cosec} 	heta = \frac{2\sqrt{3} - 2 \pm \sqrt{(2(\sqrt{3} - 1))^2 + 4 \sqrt{3} \cdot 4}}{2\sqrt{3}} $

Simplifying inside the square root:

$ (2(\sqrt{3} - 1))^2 + 4\sqrt{3} \cdot 4 = 4(3 + 1 - 2\sqrt{3}) + 16\sqrt{3} $

This simplifies to:

$ 4(4 - 2\sqrt{3}) + 16\sqrt{3} = 16 - 8\sqrt{3} + 16\sqrt{3} = 16 + 8\sqrt{3} $

Therefore, the quadratic gives:

$ \operatorname{cosec} 	heta = \frac{2\sqrt{3} - 2 \pm 2(\sqrt{3} + 1)}{2\sqrt{3}} $

By solving, we find:

$ \operatorname{cosec} 	heta = \frac{-2}{\sqrt{3}}, 2 $

Thus, for each value of the cosecant, $	heta$ can take several values that fall within the given interval $[- \frac{7\pi}{6}, \frac{4\pi}{3}]$:

$ 	heta = \frac{-7\pi}{6}, \frac{-2\pi}{3}, \frac{-\pi}{3}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{4\pi}{3} $

These are all the possible solutions for $	heta$ within the defined range.

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