$\cos 3x>\dfrac{1}{2}$ $x\in\left(-\dfrac{\pi}{6}+2\pi n;\;\dfrac{\pi}{6}+2\pi n\right),\;n\in\Z$ $x\in\left(-\dfrac{\pi}{6}+\dfrac{2\pi n}{3};\;\dfrac{\pi}{6}+\dfrac{2\pi n}{3}\right),\;n\in\Z$ $x\in\left(-\dfrac{\pi}{9}+\dfrac{2\pi n}{3};\;\dfrac{\pi}{9}+\dfrac{2\pi n}{3}\right),\;n\in\Z$ $x\in\left(\dfrac{\pi}{9}+\dfrac{2\pi n}{3};\;\dfrac{5\pi}{9}+\dfrac{2\pi n}{3}\right),\;n\in\Z$

Задание

Реши неравенство.

$\cos 3x\gt \dfrac{1}{2}$

Выбери верный вариант.

$x\in\left(-\dfrac{\pi}{6}+2\pi n;\;\dfrac{\pi}{6}+2\pi n\right),\;n\in\Z$
$x\in\left(-\dfrac{\pi}{6}+\dfrac{2\pi n}{3};\;\dfrac{\pi}{6}+\dfrac{2\pi n}{3}\right),\;n\in\Z$
$x\in\left(-\dfrac{\pi}{9}+\dfrac{2\pi n}{3};\;\dfrac{\pi}{9}+\dfrac{2\pi n}{3}\right),\;n\in\Z$
$x\in\left(\dfrac{\pi}{9}+\dfrac{2\pi n}{3};\;\dfrac{5\pi}{9}+\dfrac{2\pi n}{3}\right),\;n\in\Z$