Question

решите плиз 1-sin2x=2sinx-cosx​

Answer 1

Ответ:

$1 - \sin(2x) = 2 \sin(x) - \cos(x) \\ 1 - 2 \sin(x) \cos(x) - 2 \sin(x) + \cos(x) = 0 \\ - 2 \sin(x) ( \cos(x) + 1) + (1 + \cos(x) ) = 0 \\ (1 + \cos(x)) (1 - 2 \sin(x)) = 0 \\ \\ 1 + \cos(x) = 0 \\ \cos(x) = - 1 \\ x1 = \pi + 2\pi \: n \\ \\ 1 - 2 \sin(x) = 0 \\ \sin(x) = \frac{1}{2} \\ x2 = \frac{\pi}{6} +2 \pi \: n \\ x3 = \frac{5\pi}{6} + 2\pi \: n$

n принадлежит Z.