Question

sin2x+cos2x • sinx =0

Answer 1

Ответ:

$\sin(2x) + \cos(2x) \times \sin(x) = 0 \\ 2 \sin(x) \cos(x) + \sin(x) \cos(2x) = 0 \\ \sin(x) (2 \cos(x) + \cos(2x) ) = 0 \\ \\ \sin(x) = 0 \\ x_1 = \pi \: n \\ \\ 2 \cos(x) + \cos(2x) = 0 \\ 2 \cos(x) + 2 \cos {}^{2} (x) - 1 = 0 \\ 2 \cos {}^{2} (x) + 2 \cos(x) - 1 = 0 \\ \\ \cos(x) = t \\ \\ 2t {}^{2} + 2 t - 1 = 0\\ D = 4 + 12 = 16\\ t_1 = \frac{ - 2 + 4}{4} = \frac{1}{2} \\ t_2 = - \frac{6}{4} = - 1.5 \\ \\ \cos(x) = \frac{1}{2} \\ x_2 = \pm \frac{\pi}{3} + 2\pi \: n \\ \\ \cos(x) = - 1.5$

нет корней

Ответ:

$x_1 = \pi \: n \\ x_2 = \pm \frac{\pi}{3} + 2 \pi \: n$

n принадлежит Z