$\cos 5x<\dfrac{\sqrt{2}}{2}$ $x\in\left(-\dfrac{\pi}{20}+\dfrac{2\pi n}{5};\;\dfrac{\pi}{20}+\dfrac{2\pi n}{5}\right),\;n\in\Z$ $x\in\left(-\dfrac{\pi}{20}+2\pi n;\;\dfrac{\pi}{20}+2\pi n\right),\;n\in\Z$ $x\in\left(\dfrac{\pi}{4}+2\pi n;\;\dfrac{7\pi}{4}+2\pi n\right),\;n\in\Z$ $x\in\left(\dfrac{\pi}{20}+\dfrac{2\pi n}{5};\;\dfrac{7\pi}{20}+\dfrac{2\pi n}{5}\right),\;n\in\Z$

Задание

Реши неравенство.

$\cos 5x<\dfrac{\sqrt{2}}{2}$

Выбери верный вариант.

$x\in\left(-\dfrac{\pi}{20}+\dfrac{2\pi n}{5};\;\dfrac{\pi}{20}+\dfrac{2\pi n}{5}\right),\;n\in\Z$
$x\in\left(-\dfrac{\pi}{20}+2\pi n;\;\dfrac{\pi}{20}+2\pi n\right),\;n\in\Z$
$x\in\left(\dfrac{\pi}{4}+2\pi n;\;\dfrac{7\pi}{4}+2\pi n\right),\;n\in\Z$
$x\in\left(\dfrac{\pi}{20}+\dfrac{2\pi n}{5};\;\dfrac{7\pi}{20}+\dfrac{2\pi n}{5}\right),\;n\in\Z$