Задание

Основано на упр. 1 для самостоят.работы, стр. 6

Выбери верные ответы

Найди область определения функции:

y=\sqrt{1-\tg^{2}x};

2\pi n\lt x\lt \dfrac{\pi}{2} + 2\pi n, n\in \Z.

-\cfrac{\pi}{4}+\pi n\le x\le \cfrac{\pi}{4} + \pi n, n\in \Z.

-\cfrac{\pi}{4}+\pi n\ge x\ge \cfrac{\pi}{4} + \pi n, n\in \Z.

y=\sqrt{\ctg^{2}x-3};

\pi n\lt x\le \cfrac{\pi}{6} + \pi n, \space \cfrac{5\pi}{6}+\pi n\le x\lt \pi(1+n), n\in \Z.

-\cfrac{\pi}{6}+\pi n\ge x\ge \cfrac{\pi}{6} + \pi n, n\in \Z.

2\pi n\lt x\lt \cfrac{\pi}{4} + 4\pi n, n\in \Z.

y=\sqrt{3\sin^{2}x-\cos^{2}x};

-\cfrac{\pi}{6}+\pi n\ge x\ge \cfrac{\pi}{6} + \pi n, n\in \Z.

2\pi n\lt x\lt \cfrac{\pi}{6} + 5\pi n, n\in \Z.

\cfrac{\pi}{6} + \pi n\le x\le \cfrac{5\pi}{6} + \pi n, n\in \Z.

y=\sqrt{\sin^{2}x-3\cos^{2}x};

\cfrac{\pi}{3}+\pi n\ge x\ge \cfrac{2\pi}{3} + \pi n, n\in \Z.

\cfrac{\pi}{3} + \pi n\le x\le \cfrac{2\pi}{3} + \pi n, n\in \Z.

2\pi n\lt x\lt \cfrac{\pi}{3} + 2\pi n, n\in \Z.

y=\sqrt{\cos2x-\sin x};

\cfrac{5\pi}{6} + 2\pi n\le x\le \cfrac{13\pi}{6} + 2\pi n, n\in \Z.

\cfrac{5\pi}{6}+\pi n\ge x\ge \cfrac{13\pi}{6} + \pi n, n\in \Z.

5\pi n\lt x\lt \cfrac{13\pi}{6} + 2\pi n, n\in \Z.

y=\sqrt{3\cos2x+7\cos x};

-\arccos\cfrac{1}{3} + 2\pi n\le x\le \arccos\cfrac{1}{3} + 2\pi n, n\in \Z.

\cfrac{1}{3}+2\pi n\ge x\ge \cfrac{13\pi}{6} + \pi n, n\in \Z.

\arccos\cfrac{1}{3} +\pi n\lt x\lt \arccos\cfrac{1}{6} + 2\pi n, n\in \Z.

y=\sqrt{1-4\sin x-4\cos^{2}x};

-\cfrac{5}{6} + 2\pi n\le x\le -\cfrac{\pi}{6} + 2\pi n, n\in \Z.

\cfrac{1}{6}+2\pi n\ge x\ge \cfrac{5\pi}{6} + \pi n, n\in \Z.

-\cfrac{5\pi}{6} +2\pi n\le x\le -\cfrac{\pi}{6} + 2\pi n, n\in \Z.

y=\sqrt{2\sin 2x-2\sin^{2}x-1};

-\arctg\cfrac{1}{3} + 2\pi n\le x\le -\cfrac{\pi}{4} + 2\pi n, n\in \Z.

\arctg\,\cfrac{1}{3}+\pi n\le x\le \cfrac{\pi}{4} + \pi n, n\in \Z.

\arccos\cfrac{\pi}{4} +2\pi n\le x\le -\cfrac{\pi}{6} + 2\pi n, n\in \Z.