Упр.35.4 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

1) ctg(x+?/6)?v3; 4) tg(x/3+?/4) < v3/3; 2) cos(x/2+?/3) < -v2/2; 5) cos(x-?/6)?1/2; 3) 2sin(2?/3-x) < 1; 6) sin(4x+?/5)?-v3/2.

1. $$\ctg\left(x+\frac{\pi}{6}\right)\ge \sqrt{3}.$$

   Для функции $$\ctg t$$ имеем:

   $$\pi n<t\le \frac{\pi}{6}+\pi n,\quad n\in\mathbb Z,$$
   где $$t=x+\frac{\pi}{6}.$$

   Тогда
   $$ \pi n<x+\frac{\pi}{6}\le \frac{\pi}{6}+\pi n, $$
   откуда
   $$ -\frac{\pi}{6}+\pi n<x\le \pi n. $$

2. $$\cos\left(\frac{x}{2}+\frac{\pi}{3}\right)<-\frac{\sqrt{2}}{2}.$$

   Пусть $$t=\frac{x}{2}+\frac{\pi}{3}.$$ Тогда

   $$ \frac{3\pi}{4}+2\pi n<t<\frac{5\pi}{4}+2\pi n,\quad n\in\mathbb Z. $$

   Следовательно,
   $$ \frac{3\pi}{4}+2\pi n<\frac{x}{2}+\frac{\pi}{3}<\frac{5\pi}{4}+2\pi n, $$
   $$ \frac{5\pi}{12}+2\pi n<\frac{x}{2}<\frac{11\pi}{12}+2\pi n, $$
   $$ \frac{5\pi}{6}+4\pi n<x<\frac{11\pi}{6}+4\pi n. $$

3. $$2\sin\left(\frac{2\pi}{3}-x\right)<1.$$

   Получаем

   $$ \sin\left(\frac{2\pi}{3}-x\right)<\frac12. $$

   Это равносильно неравенству

   $$ \sin\left(x-\frac{2\pi}{3}\right)>-\frac12. $$

   Тогда

   $$ -\frac{\pi}{6}+2\pi n<x-\frac{2\pi}{3}<\frac{7\pi}{6}+2\pi n,\quad n\in\mathbb Z. $$

   Прибавляя $$\frac{2\pi}{3}$$, получаем
   $$ \frac{\pi}{2}+2\pi n<x<\frac{11\pi}{6}+2\pi n. $$

4. $$\tg\left(\frac{x}{3}+\frac{\pi}{4}\right)<\frac{\sqrt{3}}{3}.$$

   Пусть $$t=\frac{x}{3}+\frac{\pi}{4}.$$ Тогда

   $$ -\frac{\pi}{2}+\pi n<t<\frac{\pi}{6}+\pi n,\quad n\in\mathbb Z. $$

   Значит,
   $$ -\frac{\pi}{2}+\pi n<\frac{x}{3}+\frac{\pi}{4}<\frac{\pi}{6}+\pi n, $$
   $$ -\frac{3\pi}{4}+\pi n<\frac{x}{3}<-\frac{\pi}{12}+\pi n, $$
   $$ -\frac{9\pi}{4}+3\pi n<x<-\frac{\pi}{4}+3\pi n. $$

5. $$\cos\left(x-\frac{\pi}{6}\right)\ge \frac12.$$

   Для косинуса имеем

   $$ -\frac{\pi}{3}+2\pi n\le x-\frac{\pi}{6}\le \frac{\pi}{3}+2\pi n,\quad n\in\mathbb Z. $$

   Тогда
   $$ -\frac{\pi}{6}+2\pi n\le x\le \frac{\pi}{2}+2\pi n. $$

6. $$\sin\left(4x+\frac{\pi}{5}\right)\le -\frac{\sqrt{3}}{2}.$$

   Пусть $$t=4x+\frac{\pi}{5}.$$ Тогда

   $$ -\frac{2\pi}{3}+2\pi n\le t\le -\frac{\pi}{3}+2\pi n,\quad n\in\mathbb Z. $$

   Следовательно,
   $$ -\frac{2\pi}{3}+2\pi n\le 4x+\frac{\pi}{5}\le -\frac{\pi}{3}+2\pi n, $$
   $$ -\frac{13\pi}{15}+2\pi n\le 4x\le -\frac{8\pi}{15}+2\pi n, $$
   $$ -\frac{13\pi}{60}+\frac{\pi n}{2}\le x\le -\frac{2\pi}{15}+\frac{\pi n}{2}. $$

Ответ

1. $$-\frac{\pi}{6}+\pi n<x\le \pi n,\quad n\in\mathbb Z.$$
2. $$\frac{5\pi}{6}+4\pi n<x<\frac{11\pi}{6}+4\pi n,\quad n\in\mathbb Z.$$
3. $$\frac{\pi}{2}+2\pi n<x<\frac{11\pi}{6}+2\pi n,\quad n\in\mathbb Z.$$
4. $$-\frac{9\pi}{4}+3\pi n<x<-\frac{\pi}{4}+3\pi n,\quad n\in\mathbb Z.$$
5. $$-\frac{\pi}{6}+2\pi n\le x\le \frac{\pi}{2}+2\pi n,\quad n\in\mathbb Z.$$
6. $$-\frac{13\pi}{60}+\frac{\pi n}{2}\le x\le -\frac{2\pi}{15}+\frac{\pi n}{2},\quad n\in\mathbb Z.$$