Предмет: Алгебра
Автор: Аноним
Вопрос задан 2 года назад

Помогите решить тригонометрические уравнения ,надо решить 10 штук пожалуйста очень надо(

Приложения:

Ответы

Ответ дал: Miroslava227

Ответ:

$\sin(2x + \frac{\pi}{3} ) = - \frac{1}{2} \\ \\ 2x_1 + \frac{\pi}{3} = - \frac{\pi}{6} + 2 \pi \: n \\ 2x_1 = - \frac{\pi}{2} + 2\pi \: n \\ x_1 = - \frac{\pi}{4} + \pi \: n \\ \\ 2x_2 + \frac{\pi}{3} = - \frac{5\pi}{6} + 2\pi \: n \\ 2x2 = - \frac{7\pi}{6} \pi \: n \\ x_2 = - \frac{7\pi}{12} + \pi \: n$

$\cos( \frac{\pi}{4} - x) = \frac{ \sqrt{3} }{2} \\ \\ \frac{\pi}{4} - x_1 = \frac{\pi}{6} + 2\pi \: n \\ - x_1 = - \frac{\pi}{12} + 2\pi \: n \\ x_1 = \frac{\pi}{12} + 2\pi \: n \\ \\ \frac{\pi}{4} - x_2 = - \frac{\pi}{6} + 2\pi \: n \\ x_2 = \frac{5\pi}{12} + 2\pi \: n$

$tg( \frac{x}{4} - \frac{\pi}{5} ) = - \frac{ \sqrt{3} }{2} \\ \frac{x}{4} - \frac{\pi}{5} = - arctg( \frac{ \sqrt{3} }{2} ) + \pi \: n \\ \frac{x}{4} = \frac{\pi}{5} + acrtg( \frac{ \sqrt{3} }{2} ) + \pi \: n \\ x = \frac{4\pi}{5} + 4arctg( \frac{ \sqrt{3} }{2} ) + 4\pi \: n$

$ctg( \frac{x}{2} + \frac{\pi}{7} ) = - \sqrt{3} \\ \frac{x}{2} + \frac{\pi}{7} = - \frac{\pi}{6} + \pi \: n \\ \frac{x}{2} = - \frac{13\pi}{42} + \pi \: n \\ x = \frac{13\pi}{21} + 2\pi \: n$

$\sin(4x - \frac{\pi}{5} ) = - \frac{1}{2} \\ \\ 4x_1 - \frac{\pi}{5} = - \frac{\pi}{6} + 2\pi \: n \\ 4x_1 = \frac{\pi}{30} + 2 \pi \: n \\ x_1 = \frac{\pi}{120} + \frac{\pi \: n}{2} \\ \\ 4x_2 - \frac{\pi}{5} = - \frac{5\pi}{6} + 2\pi \: n \\ 4x_2 = - \frac{19\pi}{30} + 2\pi \: n \\ x_2 = - \frac{19\pi}{120} + \frac{\pi \: n}{2}$

$\cos( \frac{x}{4} - \frac{\pi}{5} ) = - \frac{1}{2 } \\ \frac{x_1}{4} - \frac{\pi}{5} = \frac{2\pi}{3} + 2 \pi \: n \\ \frac{x_1}{4} = \frac{13\pi}{15} + 2\pi \: n \\ x_1 = \frac{52\pi}{15} + 8 \pi \: n \\ \\ \frac{x_2}{4} - \frac{\pi}{5} = - \frac{2\pi}{3} + 2\pi \: n \\ \frac{x_2}{4} = - \frac{7\pi}{15} + 2\pi \: n \\ x_2 = - \frac{28\pi}{15} + 8\pi \: n$

$tg(2x + \frac{\pi}{6} ) = 8 \\ 2x + \frac{\pi}{6} = arctg(8) + \pi \: n \\ x = - \frac{\pi}{12} + \frac{1}{2} arctg(8) + \frac{\pi \: n}{2}$

$ctg(3x - \frac{\pi}{4} ) = \frac{ \sqrt{3} }{3} \\ 3x - \frac{\pi}{4} = \frac{\pi}{3} + \ pi \: n \\ 3x = \frac{7\pi}{12} + \pi \: n \\ x = \frac{7 \pi }{36} + \frac{\pi \: n}{3}$

11.

$tg( \frac{\pi}{3} - 4x) = - \sqrt{3} \\ \frac{\pi}{4} - 4x = - \frac{\pi}{3} + \pi \: n \\ - 4x = - \frac{7\pi}{12} + \pi \: n \\ x = \frac{\pi}{48} + \frac{ \pi \: n}{4}$