$\cos 4x<\dfrac{\sqrt{3}}{2}$ $x\in\left(\dfrac{\pi}{24}+2\pi n;\;\dfrac{11\pi}{24}+2\pi n\right),\;n\in\Z$ $x\in\left(-\dfrac{\pi}{16}+\dfrac{\pi n}{2};\;\dfrac{\pi}{16}+\dfrac{\pi n}{2}\right),\;n\in\Z$ $x\in\left(\dfrac{\pi}{24}+\dfrac{\pi n}{2};\;\dfrac{11\pi}{24}+\dfrac{\pi n}{2}\right),\;n\in\Z$ $x\in\left(\dfrac{\pi}{12}+\dfrac{2\pi n}{3};\;\dfrac{5\pi}{12}+\dfrac{2\pi n}{3}\right),\;n\in\Z$

Задание

Реши неравенство.

$\cos 4x\lt \dfrac{\sqrt{3}}{2}$

Выбери верный вариант.

$x\in\left(\dfrac{\pi}{24}+2\pi n;\;\dfrac{11\pi}{24}+2\pi n\right),\;n\in\Z$
$x\in\left(-\dfrac{\pi}{16}+\dfrac{\pi n}{2};\;\dfrac{\pi}{16}+\dfrac{\pi n}{2}\right),\;n\in\Z$
$x\in\left(\dfrac{\pi}{24}+\dfrac{\pi n}{2};\;\dfrac{11\pi}{24}+\dfrac{\pi n}{2}\right),\;n\in\Z$
$x\in\left(\dfrac{\pi}{12}+\dfrac{2\pi n}{3};\;\dfrac{5\pi}{12}+\dfrac{2\pi n}{3}\right),\;n\in\Z$