Question

Помогите пожалуйста

Answer 1

в)

$2 \sin(x) + \sin(2x) = \cos(x) + 1 \\ 2 \sin(x) + 2 \sin(x) \cos(x) = \cos(x) + 1 \\ 2 \sin(x) ( \cos(x) + 1) = \cos(x) + 1 \\ 2 \sin(x) = \frac{ \cos(x) + 1 }{ \cos(x) + 1} \\ 2 \sin(x) = 1, \: \cos(x) + 1≠0 \\ \sin(x) = \frac{1}{2} , \: x≠\pi + 2\pi n \\ x = ( - 1)^{n} \frac{\pi}{6} + \pi n, \: n \in\mathbb Z$

д)

$\cos(2x) = \sin(x) \\ 1 - 2 \sin^{2} (x) = \sin(x) \\ 2 \sin^{2} (x) + \sin(x) - 1 = 0 \\ \sin(x) = - 1 \\ \sin(x) = \frac{1}{2} \\ x = \frac{3\pi}{2} + 2\pi n \\ x = ( - 1)^{n} \frac{\pi}{6} + \pi n$

е)

$7 + \cos(2x) = 2 \cos(x) \\ 7 + 2 \cos^{2} (x) - 1 - 2 \cos(x) = 0 \\ 2 \cos^{2} (x) - 2 \cos(x) + 6 = 0 \\ \cos^{2} (x) - \cos(x) + 3 = 0 \\ \cos(x) \notin\mathbb R \\ x \notin\mathbb R$