Question

Решите уравнения
а) $cos(2x- \frac{ \pi }{6} )=1$
б) $\sqrt{2} sin \frac{x}{3} =1$

Answer 1

$\cos(2x- \frac{\pi}{6})=1 \\ 2x-\frac{\pi}{6}=2 \pi n,n \in Z \\ 2x=\frac{\pi}{6}+2\pi n,n \in Z \\ x= \frac{\pi}{18} +\pi n,n \in Z$

$\sqrt{2} \sin \frac{x}{3} =1 \\ \sin \frac{x}{3} = \frac{1}{\sqrt{2}} \\ \frac{x}{3} =(-1)^k\cdot \arcsin\frac{1}{\sqrt{2}}+\pi k,k \in Z \\ \frac{x}{3}= (-1)^k\cdot \frac{\pi}{4}+\pi k,k \in Z \\ x=(-1)^k\cdot \frac{3\pi}{4} +3\pi k,k \in Z$