Упр.11.53 ГДЗ Никольский 10 класс (Алгебра)

Совет дня! Чтобы не спалиться на списывании, сайт решак.ру рекомендует максимально изменять тексты сочинений «под себя», изменять имена собственные, города, подбирать синонимы.

1. а)

   $$2\sqrt{3}\sin^2 x+2\sin x\cos x>\sqrt{3}+1$$
   $$\sqrt{3}(1-\cos 2x)+\sin 2x>\sqrt{3}+1$$
   $$-\sqrt{3}\cos 2x+\sin 2x>1$$
   $$\frac12\sin 2x-\frac{\sqrt{3}}{2}\cos 2x>\frac12$$
   $$\sin\left(2x-\frac{\pi}{3}\right)>\frac12$$
   $$\frac{\pi}{6}+2\pi n<2x-\frac{\pi}{3}<\frac{5\pi}{6}+2\pi n,\quad n\in\mathbb Z$$
   $$\frac{\pi}{2}+2\pi n<2x<\frac{7\pi}{6}+2\pi n$$
   $$\frac{\pi}{4}+\pi n<x<\frac{7\pi}{12}+\pi n,\quad n\in\mathbb Z$$

2. б)

   $$2\sin^2 x-2\sqrt{3}\sin x\cos x>\sqrt{2}-1$$
   $$1-\cos 2x-\sqrt{3}\sin 2x>\sqrt{2}-1$$
   $$-\cos 2x-\sqrt{3}\sin 2x>\sqrt{2}-2$$
   $$\cos 2x+\sqrt{3}\sin 2x<2-\sqrt{2}$$
   $$\frac12\cos 2x+\frac{\sqrt{3}}{2}\sin 2x<\frac{2-\sqrt{2}}{2}$$
   $$\sin\left(2x+\frac{\pi}{6}\right)<\frac{2-\sqrt{2}}{2}$$
   $$-\frac{\pi}{4}+2\pi n<2x+\frac{\pi}{6}<\frac{5\pi}{4}+2\pi n,\quad n\in\mathbb Z$$
   $$-\frac{5\pi}{12}+2\pi n<2x<\frac{13\pi}{12}+2\pi n$$
   $$-\frac{5\pi}{24}+\pi n<x<\frac{13\pi}{24}+\pi n,\quad n\in\mathbb Z$$

3. в)

   $$2\sin^2 x+2\sqrt{3}\sin x\cos x<\sqrt{2}+1$$
   $$1-\cos 2x+\sqrt{3}\sin 2x<\sqrt{2}+1$$
   $$-\cos 2x+\sqrt{3}\sin 2x<\sqrt{2}$$
   $$\frac12\cos 2x-\frac{\sqrt{3}}{2}\sin 2x>-\frac{\sqrt{2}}{2}$$
   $$\sin\left(2x-\frac{\pi}{6}\right)>-\frac{\sqrt{2}}{2}$$
   $$-\frac{\pi}{4}+2\pi n<2x-\frac{\pi}{6}<\frac{5\pi}{4}+2\pi n,\quad n\in\mathbb Z$$
   $$-\frac{\pi}{12}+2\pi n<2x<\frac{17\pi}{12}+2\pi n$$
   $$-\frac{\pi}{24}+\pi n<x<\frac{17\pi}{24}+\pi n,\quad n\in\mathbb Z$$

4. г)

   $$2\sin^2 x-2\sqrt{3}\sin x\cos x<\sqrt{3}$$
   $$1-\cos 2x-\sqrt{3}\sin 2x<\sqrt{3}$$
   $$-\cos 2x-\sqrt{3}\sin 2x<\sqrt{3}-1$$
   $$\cos 2x+\sqrt{3}\sin 2x>1-\sqrt{3}$$
   $$\frac12\cos 2x+\frac{\sqrt{3}}{2}\sin 2x>\frac{1-\sqrt{3}}{2}$$
   $$\sin\left(2x+\frac{\pi}{6}\right)>\frac{1-\sqrt{3}}{2}$$
   $$-\frac{\pi}{4}+2\pi n<2x+\frac{\pi}{6}<\frac{5\pi}{4}+2\pi n,\quad n\in\mathbb Z$$
   $$-\frac{5\pi}{12}+2\pi n<2x<\frac{13\pi}{12}+2\pi n$$
   $$-\frac{5\pi}{24}+\pi n<x<\frac{13\pi}{24}+\pi n,\quad n\in\mathbb Z$$

Ответ

а) $$\frac{\pi}{4}+\pi n<x<\frac{7\pi}{12}+\pi n,\ n\in\mathbb Z$$

б) $$-\frac{5\pi}{24}+\pi n<x<\frac{13\pi}{24}+\pi n,\ n\in\mathbb Z$$

в) $$-\frac{\pi}{24}+\pi n<x<\frac{17\pi}{24}+\pi n,\ n\in\mathbb Z$$

г) $$-\frac{5\pi}{24}+\pi n<x<\frac{13\pi}{24}+\pi n,\ n\in\mathbb Z$$