Question

Тригонометрические уравнения и неравенства

Answer 1

Ответ:

$x = \frac{\pi}{18} + \frac{\pi k}{6}$; $x = \frac{\pi}{12} + \frac{\pi k}{4}$, k ∈ Z

Объяснение:

$\sqrt{3} * sin(x) - cos(x) = 2cos(7x)\\\\2sin(x-\frac{\pi}{6}) = 2cos(7x)\\\\sin(x-\frac{\pi}{6}) - sin(\frac{\pi}{2} - 7x) = 0\\\\2sin(\frac{x-\frac{\pi}{6} - \frac{\pi}{2} + 7x}{2})cos(\frac{x-\frac{\pi}{6} + \frac{\pi}{2} - 7x}{2}) = 0\\\\sin(4x - \frac{\pi}{3})cos(3x - \frac{\pi}{6}) = 0$

1) $sin(4x - \frac{\pi}{3}) = 0$

$4x - \frac{\pi}{3} = \pi k\\x = \frac{\pi}{12} + \frac{\pi k}{4}$

или

2) $cos(3x - \frac{\pi}{6}) = 0$

$3x - \frac{\pi}{6} = \frac{\pi k}{2}\\x = \frac{\pi}{18} + \frac{\pi k}{6}$

k ∈ Z