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If $\tan^{3} x + 3 > 3 \tan x + \tan^{2} x,$ then

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$x \in (n π + \frac{π}{3}, n π + \frac{π}{2}) \cup (n π - \frac{π}{3}, n π + \frac{π}{4}); n \in Z$

$x \in (n π + \frac{π}{4}, n π + \frac{π}{2}) \cup (n π - \frac{π}{4}, n π + \frac{π}{4}); n \in Z$

$x \in (n π - \frac{π}{6}, n π + \frac{π}{2}) \cup (n π - \frac{π}{4}, n π + \frac{π}{4}); n \in Z$

$x \in (n π - \frac{π}{3}, n π + \frac{π}{2}) \cup (n π - \frac{π}{4}, n π + \frac{π}{4}); n \in Z$

answer is A.

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Detailed Solution

$\tan^{3} x - \tan^{2} x + 3 - 3 \tan x > 0$

$\Rightarrow (\tan x - 1) (\tan^{2} x - 3) > 0$ $∴$ General solution is

$x \in (n π + \frac{π}{3}, n π + \frac{π}{2}) \cup (n π - \frac{π}{3}, n π + \frac{π}{4}) where n \in Z .$

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