Question

Решить уравнение: (фото)

Answer 1

Ответ:

а

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

б

$3 \sin {}^{2} (x) - 5\sin(x) - 2 = 0 \\ \\ \sin(x) = t \\ \\ 3t {}^{2} - 5 t - 2 = 0\\ D = 25 + 24 = 49\\ t_1 = \frac{5 + 7}{6} = 2 \\ t_2 = - \frac{1}{3} \\ \\ \sin(x) = 2$

нет корней

$\sin(x) = - \frac{1}{3} \\ x = {( - 1)}^{n + 1} arcsin( \frac{1}{3}) + \pi \: n$

в

${tg}^{2} x + 3tgx = 0 \\ \\ tgx = t \\ \\ t {}^{2} + 3 t = 0\\ t(t + 3) = 0 \\ t_1 = 0\\ t_2 = - 3 \\ \\ tgx = 0 \\ \sin(x) = 0 \\ x_1 = \pi \: n \\ \\ tgx = - 3 \\ x_2 = - arctg(3) + \pi \: n$

везде n принадлежит Z