Упр.18.15 ГДЗ Мордкович 10-11 класс (Алгебра)

a) sin 3x = корень(2)/2, [0; 2пи];
б) cos 3x = корень(3)/2, Г-пи; пи];
В) tg x/2 = корень(3)/3, [-Зпи; Зпи];
г) ctg 4x = -1, [О; пи].

1. $$\sin 3x=\frac{\sqrt2}{2}, \quad x\in[0;2\pi]$$

   $$3x=\frac{\pi}{4}+2\pi k \quad \text{или} \quad 3x=\frac{3\pi}{4}+2\pi k,\quad k\in\mathbb Z$$

   $$x=\frac{\pi}{12}+\frac{2\pi k}{3} \quad \text{или} \quad x=\frac{\pi}{4}+\frac{2\pi k}{3}$$

   На отрезке $$[0;2\pi]$$ получаем:

   $$\frac{\pi}{12},\ \frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{11\pi}{12},\ \frac{17\pi}{12},\ \frac{19\pi}{12}$$

2. $$\cos 3x=\frac{\sqrt3}{2}, \quad x\in[-\pi;\pi]$$

   $$3x=\pm\frac{\pi}{6}+2\pi k,\quad k\in\mathbb Z$$

   $$x=\pm\frac{\pi}{18}+\frac{2\pi k}{3}$$

   На отрезке $$[-\pi;\pi]$$ получаем:

   $$-\frac{13\pi}{18},\ -\frac{11\pi}{18},\ -\frac{\pi}{18},\ \frac{\pi}{18},\ \frac{11\pi}{18},\ \frac{13\pi}{18}$$

3. $$\tg\frac{x}{2}=\frac{\sqrt3}{3}, \quad x\in[-3\pi;3\pi]$$

   $$\frac{x}{2}=\frac{\pi}{6}+\pi k,\quad k\in\mathbb Z$$

   $$x=\frac{\pi}{3}+2\pi k$$

   На отрезке $$[-3\pi;3\pi]$$ получаем:

   $$-\frac{5\pi}{3},\ \frac{\pi}{3},\ \frac{7\pi}{3}$$

4. $$\ctg 4x=-1, \quad x\in[0;\pi]$$

   $$4x=-\frac{\pi}{4}+\pi k,\quad k\in\mathbb Z$$

   $$x=-\frac{\pi}{16}+\frac{\pi k}{4}$$

   На отрезке $$[0;\pi]$$ получаем:

   $$\frac{3\pi}{16},\ \frac{7\pi}{16},\ \frac{11\pi}{16},\ \frac{15\pi}{16}$$

Ответ

а) $$\frac{\pi}{12},\ \frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{11\pi}{12},\ \frac{17\pi}{12},\ \frac{19\pi}{12}$$;
б) $$-\frac{13\pi}{18},\ -\frac{11\pi}{18},\ -\frac{\pi}{18},\ \frac{\pi}{18},\ \frac{11\pi}{18},\ \frac{13\pi}{18}$$;
в) $$-\frac{5\pi}{3},\ \frac{\pi}{3},\ \frac{7\pi}{3}$$;
г) $$\frac{3\pi}{16},\ \frac{7\pi}{16},\ \frac{11\pi}{16},\ \frac{15\pi}{16}$$.