, 2 ВАРИАНТ НАДО РЕШИТЬ НЕ КОНТРОЛЬНАЯ А ДЗ НА ДОСКЕ НАПИСАЛИ, ПОЧЕМУ БАНЯТ ТО

Ответ:

Chundokova

15.05.2021 16:38

$\sqrt{2} \sin(x) - 1 = 0 \\ \sin(x) = \frac{1}{ \sqrt{2} } = \frac{ \sqrt{2} }{2} \\ x_1 = \frac{\pi}{4} + 2 \pi \: n \\ x_2 = \frac{3\pi}{4} + 2 \pi \: n$

$tg \frac{x}{2} = \sqrt{3} \\ \frac{x}{2} = \frac{\pi}{3} + \pi \: n \\ x = \frac{2\pi}{3} + 2\pi \: n$

$10 \cos {}^{2} (x) + 3 \cos(x) - 1 = 0 \\ \\ \cos(x) = t \\ \\ 10t {}^{2} + 3 t - 1 = 0 \\ D = 9 + 40 = 49\\ t_1 = \frac{ - 3 + 7}{20} = 0.2 \\ t_2 = - \frac{1}{2} \\ \\ \cos(x) =0.2 \\ x_1 = \pm \: arccos(0.2) + \pi \: n \\ \\ \cos(x) = - \frac{1}{2} \\ x_2 = \pm \frac{\pi}{3} + 2 \pi \: n$

$\sin {}^{2} (x) + 2 \cos {}^{2} (x) - 5 \cos(x) - 7 = 0 \\ 1 - \cos {}^{2} (x) + 2\cos {}^{2} (x) - 5 \cos(x) - 7 = 0 \\ \cos {}^{2} (x) - 5 \cos(x) - 6 = 0 \\ \\ \cos(x) = t \\ \\ t {}^{2} - 5 t - 6 = 0\\ D = 25 + 24 = 49\\ t_1= \frac{5 + 7}{2} = 6 \\ t_2 = - 1 \\ \\ \cos(x) = 6 \\ \text{нет корней} \\ \\ \cos(x) = - 1 \\ x = \pi + 2\pi \: n$

$\sin(x) - \sqrt{3} \cos(x) = 0 \\ \sin(x) = \sqrt{3} \cos(x) \\ | \div \cos(x) \ne0 \\ tgx = \sqrt{3} \\ x = \frac{\pi}{3} + \pi \: n$

$\sin {}^{2} (x) + 6 \cos {}^{2} (x) + 7 \sin( {x}^{} ) \cos( {x}^{} ) = 0 \\ | \div \cos {}^{2} (x) \ne0 \\ {tg}^{2} x + 7tgx + 6 = 0 \\ \\ tgx = t \\ \\ t {}^{2} + 7t + 6 = 0\\ D = 49 - 24 = 25 \\ t_1 = \frac{ - 7 + 5}{2} = - 1 \\ t_2 = - 6 \\ \\ tgx = - 1 \\ x_1 = - \frac{\pi}{4} + \pi \: n \\ \\ tgx = - 6 \\ x_2 = - arctg(6) + \pi \: n$