Question

Алгебра. Задание 252.

Answer 1

решим применяя формулы приведения

$1) \sin(5\pi + x) = 1 \\ \sin(\pi + 4\pi + x) = 1 \\ \sin(\pi + x) = 1 \\ - \sin \: x = 1 \\ sinx = - 1 \\ x = - \frac{\pi}{2} + 2\pi \: n ,n∈Z$

$2) \cos(x - 3\pi) = 0 \\ \cos(x + \pi - 4\pi) = 0 \\ \cos(x + \pi) = 0 \\ - \cos \: x = 0 \\ cosx = 0 \\ x = \frac{\pi}{2} + \pi \: n,n∈Z$

$3) \cos( \frac{5\pi}{2} + x ) = - 1 \\ \cos( \frac{\pi}{2} + 2\pi + x ) = - 1 \\ \cos( \frac{\pi}{2} + x ) = - 1 \\ - \sin \: x = - 1 \\ sinx = 1 \\ x = \frac{\pi}{2} + 2\pi \: n ,n∈Z$

$4) \sin( \frac{9\pi}{2} + x ) = - 1 \\ \sin( \frac{\pi}{2} + 4\pi + x ) = - 1 \\ \sin( \frac{\pi}{2} + x ) = - 1 \\ cosx = - 1 \\ x = \pi + 2\pi \: n,n∈Z$