Question

$\sqrt{3} \tan(?) (x + \frac{\pi}{6} ) > 1$
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Answer 1

Ответ:

$\pi n < x < \frac{\pi}{3} + \pi n$

Объяснение:

$\sqrt{3} tan(x + \frac{\pi}{6} ) > 1\\\\tan(x + \frac{\pi}{6} ) > \frac{1}{\sqrt{3}} \\\\tan(x + \frac{\pi}{6} ) > tan \frac{\pi}{6} \\\\\frac{\pi }{6} + \pi n < x + \frac{\pi}{6} < \frac{\pi}{2} + \pi n \\\\\frac{\pi }{6} + \pi n - \frac{\pi}{6} < x < \frac{\pi}{2} + \pi n - \frac{\pi}{6}\\\\\pi n < x < \frac{\pi}{3} + \pi n$