The number of solutions of the inequality $E = 2^{1/\sin^{2}\alpha_{2}} \bullet 3^{1/\sin^{2}\alpha_{3}}\ldots n^{1/\sin^{2}\alpha_{n}}

Question

The number of solutions of the inequality $E = 2^{1/\sin^{2}\alpha_{2}} \bullet 3^{1/\sin^{2}\alpha_{3}}\ldots n^{1/\sin^{2}\alpha_{n}} < n\ !$ Where $\alpha_{i} \in ( - \pi,2\pi)$ for $i = 2,\ 3,\ldots,n\$is

Dr. Abubacker · Accepted Answer

(c)
Given inequality holds only, if
$\sin^{2}{\alpha_{i} = 1\ or\ \alpha_{i} = \pm \frac{\pi}{2},\frac{3\pi}{2};\ \ \ (i = 2,3,\ldots,n)}$
$\Rightarrow \$Number of solutions$= 3 	imes 3 	imes 3 	imes \ldots 	imes (n - 1)$times
$= 3^{n - 1}$

Q. The number of solutions of the inequality $E = 2^{1/\sin^{2}\alpha_{2}} \bullet 3^{1/\sin^{2}\alpha_{3}}\ldots n^{1/\sin^{2}\alpha_{n}} < n\ !$ Where $\alpha_{i} \in ( - \pi,2\pi)$ for $i = 2,\ 3,\ldots,n\$is

Explanation

Reviewed by Dr. Abubacker

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