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Question

(A) Given $\cot^3	heta = -3\sqrt{3} = -(\sqrt{3})^3$,so $\cot	heta = -\sqrt{3}$.This implies $	an	heta = -\frac{1}{\sqrt{3}}$,which means $	heta

Vedclass · Accepted Answer

$2n\pi - \frac{\pi}{6}$

Vedclass · Answer

(A) Given $\cot^3	heta = -3\sqrt{3} = -(\sqrt{3})^3$,so $\cot	heta = -\sqrt{3}$.This implies $	an	heta = -\frac{1}{\sqrt{3}}$,which means $	heta = n\pi - \frac{\pi}{6}$.Given $\csc^5	heta = -32 = (-2)^5$,so $\csc	heta = -2$.This implies $\sin	heta = -\frac{1}{2}$,which means $	heta = n\pi + (-1)^n(-\frac{\pi}{6}) = n\pi - (-1)^n\frac{\pi}{6}$.For $\cot	heta = -\sqrt{3}$ and $\sin	heta = -\frac{1}{2}$,$	heta$ must be in the fourth quadrant.In the fourth quadrant,the general solution is $	heta = 2n\pi - \frac{\pi}{6}$.

The general value of $\theta$ that satisfies both the equations $\cot^3\theta + 3\sqrt{3} = 0$ and $\csc^5\theta + 32 = 0$ is $(n \in I)$.

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